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337. A sheet of paper was torn into 5 pieces, some of these pieces were torn into 5 parts, some of these new parts were torn again into 5 parts, and so on. Can this method result in 1994 pieces of paper? What about 1997? | $\triangle$ Each time a sheet or a piece of paper is torn into 5 parts, the total number of pieces increases by 4. Therefore, the number of paper pieces at each step can only be of the form $4k+1 (k \in N)$. This expression is the invariant.
Since $1994 \neq 4k+1 (k \in N)$, it is impossible to obtain 1994 pieces, whi... | 1997 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 36,262 |
339. Three grasshoppers are playing leapfrog: if a grasshopper from point $A$ jumps over a grasshopper located at point $B$, it will land at point $C$, which is symmetric to point $A$ relative to point $B$. Initially, the grasshoppers occupy three vertices of a square. Can they, by playing leapfrog, reach the fourth ve... | $\triangle$ Let's introduce a coordinate system on the plane such that the three vertices of the square, where the grasshoppers are located, have coordinates $(0 ; 0),(0 ; 1)$, and $(1 ; 0)$. With the specified jumps, each of the coordinates of the grasshoppers either remains unchanged (for example, when jumping along ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 36,263 |
342. Each natural number from 1 to 50000 is replaced by the number equal to the sum of its digits. The same operation is performed with the resulting numbers, and this is repeated until all numbers become single-digit. How many times will each of the integers from 0 to 8 appear among these single-digit numbers? | The specified single-digit numbers in sequential order are:
$$
1,2,3,4,5,6,7,8,0,1,2,3,4,5,6,7,8,0, \ldots
$$
This pattern continues. Indeed, when a natural number is replaced by the sum of its digits, the remainder when divided by 9 remains unchanged (prove this!). Therefore, when transitioning from each natural num... | 1,2,3,4,5-5556 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 36,264 |
346. A circle is divided into 6 equal sectors, in which the numbers $0,1,2,0,2,1$ (in that order) are placed. It is allowed to add the same number to two adjacent numbers in one move. Is it possible to achieve, in several such moves, that all 6 numbers in the sectors are equal? | $\triangle$ Let at some step the numbers in the sectors are in sequential order $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$. We form the following sum:
$$
S=a_{1}-a_{2}+a_{3}-a_{4}+a_{5}-a_{6}
$$
After each move, this sum does not change, as each of the differences $a_{1}-a_{2}, a_{3}-a_{4}, a_{5}-a_{6}$ remains the s... | cannot | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,265 |
350. Among 18 coins, one is counterfeit. Genuine coins weigh the same, while the counterfeit coin differs in weight from the genuine ones. What is the minimum number of weighings on correct balance scales without weights needed to determine whether the counterfeit coin is lighter or heavier than the genuine ones? (Ther... | $\triangle$ Let's number the coins. Divide the set of coins into three piles, with 6 coins in each.
For the first weighing, place all the coins from the first pile on one scale pan, and all the coins from the second pile on the other. There are two possible cases.
1) Suppose the scales balance during this weighing. T... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,266 |
353. How can you weigh a load on a balance scale with weights if the weights are correct, but the scale is incorrect? | $\triangle$ Balance the load with weights. Then remove the load, leaving the weights on the other scale pan, and replace it with a new set of weights so that the scales are in balance again. The load weighs as much as this set. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,267 |
356. There are four packages of different masses and accurate balance scales without weights. How can you arrange the packages in order of increasing mass in five weighings? | $\triangle$ Let's number the packages. The masses of the first, second, third, and fourth packages will be denoted by $m_{1}, m_{2}, m_{3}$, and $m_{4}$, respectively.
For the first weighing, we place, for example, the first package on one pan and the second on the other; for the second weighing, the first package on ... | m_{2}<m_{1}<m_{4}<m_{3} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,268 |
358. On the street, four girls - Anya, Vanya, Galina, and Nina - are standing in a circle, talking. The girl in the green dress (not Anya and not Vanya) is standing between the girl in the blue dress and Nina. The girl in the white dress is standing between the girl in the pink dress and Vanya. What dress is each of th... | $\triangle$ This problem is similar to the correspondence problems between two sets that we considered in § 2-3.
Let's make a drawing (Fig. 42). Denote the girls in green and blue dresses, as well as Nina, with points.
Since the girl in the green dress is not Anya, not Vanya, and not Nina, her name is Galya.
The gir... | Anya-inthewhitedress,Vanya-inthebluedress,Galya-inthegreendress,Nina-inthepinkdress | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,269 |
364. Once I decided to take a ride on a chairlift. At some point, I noticed that the chair coming towards me had the number 95, and the next one had the number 0, followed by 1, 2, and so on. I looked at the number on my chair; it turned out to be 66. Have I passed the halfway point? At which chair will I pass the half... | $\triangle$ Total number of chairs is 96, and half the journey consists of 48 chairs. I will be in the middle of the cable car when the number of chairs ahead and behind me is the same. For this to happen, the number of the oncoming chair must be equal to $66-48=18$. Since the encounter with chair number 18 is ahead of... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,270 |
368. In each cell of a $3 \times 3$ square, a number is written such that the sum of any three numbers in the same row, the same column, and on each of the two diagonals is the same. In the middle of the top row, there is a 1, and in the corner cells of the bottom row, there are 2 and 3. Find the numbers written in the... | $\triangle$ Let's introduce notations for the unknown numbers of the square (Fig. 43).
Since by the condition
$$
1+d+k=2+k+3
$$
then $d=4$. Since
$$
a+1+b=2+d+b
$$
where $d$ is already known, then $a=5$. Further, since
$$
5+4+3=2+4+b
$$
then $b=6$. Similarly, we find the numbers $c, e$ and $k$.
Answer: $a=5, b=... | =5,b=6,=5,=4,e=3,k=7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,271 |
370. In each cell of a $15 \times 15$ square, one of the first 15 natural numbers is written. Two identical numbers cannot be in the same row or column, and in cells symmetric with respect to one of the two main diagonals, the same numbers are written. Can the numbers written on this main diagonal be equal? | $\triangle$ In each row and each column of the square, all natural numbers from 1 to 15 are placed, and each number appears only once. Therefore, each of these numbers appears in the square exactly 15 times.
Take, for example, the number 1. Symmetrically relative to the main diagonal of the square, only an even number... | theycannot | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,272 |
375*. In the cells of a $4 \times 4$ table, 6 asterisks are written, one in each cell. Prove that it is always possible to cross out two rows and two columns of the table so that all asterisks are crossed out. | $\triangle$ Since there are 6 asterisks and the table has 4 rows, there will be a row with more than one asterisk. Let's consider two cases.
1) Suppose in some row of the table there are at least 3 asterisks. Cross out this row. If now there are 3 asterisks left unstruck, then cross out another row and two columns in ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,274 |
378. In the prize draw for the opening of the football season in a single round-robin, five teams A, B, C, D, and E participated and took places in the same order, starting from first. Team A did not have a single draw, Team B did not lose any matches, and Team C did not win any matches. All teams scored a different nu... | $\triangle$ What was the outcome of the meeting between teams A and B? Since team A did not draw a single game, and team B did not lose a single game, team B won against A. The rest of the games were won by team A, because otherwise, it would have scored fewer than 6 points, i.e., no more than 4 points out of 8 possibl... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,275 |
386. There are 7 middle schools in the district. The district has allocated 20 computers. Prove that in any distribution of them among the schools, there will be two schools that receive the same number of computers (which could be zero). | $\triangle$ Here, the role of boxes is played by schools, and the role of objects is played by computers. Therefore, $n=7, \frac{n(n-1)}{2}=21$.
Since the number of computers is less than 21, according to the principle of insufficiency, there will always be two schools that receive the same number of computers. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,278 |
390. There are 200 candies. What is the minimum number of schoolchildren to whom these candies can be distributed so that among them, there will always be at least two who receive the same number of candies (which could be zero)? | $\triangle$ Let's find the minimum $n$ such that $200400$. There is no need to solve this quadratic inequality in the usual way. Since the product $n(n-1)$ increases with $n$, the minimum $n$ satisfying the inequality can be found by trial and error: $n=21$.
This means that if there are 21 schoolchildren, there will a... | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,279 |
392. On 5 shelves of a bookcase, 160 books are arranged, and on one of them, there are 3 books. Prove that there is a shelf with at least 40 books. | $\triangle$ In proving this statement, we use not the principle of insufficiency itself, but the idea of insufficiency.
Assume the opposite: on each shelf of the cabinet, there are no more than 39 books. Then on all 5 shelves together, there are no more than $3+4 \cdot 39=159$ books, while in reality, there are 160. W... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,280 |
395. The plane is painted in two colors - white and black, and there are both white and black points. Prove that there will always be two points of the same color at a distance of 1 from each other. | $\triangle$ Let's take an equilateral triangle with a side length of 1. Out of its three vertices, at least two are of the same color. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,281 |
398. The plane is colored in two colors - white and black, and there are both white and black points. Prove that there always exists an isosceles triangle with vertices of the same color. | $\triangle$ Let's take a regular pentagon. Among its five vertices, at least three are of the same color, and they form the vertices of an isosceles triangle. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,282 |
400. A plane is painted in three colors. Prove that there will always be two points of the same color, the distance between which is 1. | $\triangle$ Suppose the opposite: any two points that are 1 unit apart are colored differently.
Consider an equilateral triangle $ABC$ with a side length of 1. According to our assumption, all its vertices are of different colors. Now, consider the point $A_{1}$, which is symmetric to point $A$ with respect to the lin... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,283 |
404. The bottom of a rectangular box was tiled with rectangular tiles of size $2 \times 2$ and $1 \times 4$. The tiles were spilled out of the box and in the process, a tile of size $2 \times 2$ was lost. Instead of it, a tile of size $1 \times 4$ was found. Can the bottom of the box be retiled in this case? | $\triangle$ Let's divide the bottom of the box into $1 \times 1$ squares and color it in two colors as shown in Figure 47. Each tile of size $2 \times 2$ covers exactly one square colored black, so if a $2 \times 2$ tile is lost, only one black square will remain uncovered. However, a tile of size $1 \times 4$ covers e... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,284 |
409. On a plane, $n$ points are connected by segments such that each point is connected to any other (possibly through intermediate points) and there are no two points that are connected by two different paths. Prove that the total number of segments is $n-1$.
(This is one of the theorems of graph theory - the theorem... | $\triangle$ Obviously, it is assumed that $n \geq 2$.
1) For $n=2$, the statement is true: if there are two points, there is only one segment connecting them.
2) Assume that the statement is true for a natural number $n=k$, where $k \geq 2$: if there are $k$ points, there are exactly $k-1$ segments connecting them. We... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,285 |
411. In a round-robin chess tournament, there were no draws. For how many participants can it be that all chess players will score the same number of points? | Let's denote the number of participants in the tournament by $n$. Together, they scored $\frac{n(n-1)}{2}$ points, and therefore, each of them scored -
$$
\frac{n(n-1)}{2} : n = \frac{n-1}{2} \text{.}
$$
From this, $n$ is odd.
So far, we have proven that if all participants scored the same number of points, then the... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,286 |
415. Five numbers were written on the board. Then these numbers were erased and their pairwise sums were written:
$$
4,8,10,12,14,18,19,21,25,29
$$
What five numbers were written on the board?
 \cdot 1+\left(a_{2}-a_{3}\right) \cdot 2+\left(a_{3}-... | a_{1}+a_{2}+\ldots+a_{10} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,290 |
444. Eight hockey teams play against each other in a round-robin tournament to determine the final four. What is the minimum number of points that guarantees a team's advancement to the final four? | $\triangle$ All eight teams together will score $\frac{8 \cdot 7}{2} \cdot 2=56$ points. Therefore, 7, 8, and even 9 points do not guarantee a team's advancement to the final four.
What about 10 points? Imagine that there are five teams that played all their matches against each other to a draw, while the other three ... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,291 |
446. Among several jugs, there are two jugs of different shapes and two jugs of different colors. Prove that among them, there will be two jugs that are both of different shapes and different colors. | $\triangle$ Let's take two jugs of different shapes.
If they are of different colors, then they are the ones we are looking for.
If they are of the same color, then we will take a third jug that differs in color from them (by condition, it exists). Its shape differs from the shape of at least one of the two first jug... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 36,292 |
449. On a shelf, a ten-volume collection of Ilf and Petrov's works stands in a random order. A librarian can take any volume from the shelf and place it in the fifth position, counting from the left. Will the librarian be able to arrange all the volumes in ascending order of their numbers using several such operations? | $\triangle$ Let's take volume No. 1 and place it in the fifth position. In addition, all volumes that were to its left will also be placed in sequential order, one by one, in the fifth position. As a result, the first volume will end up in the first position from the left.
Now let's place volume No. 10 in the fifth po... | willbeableto | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,293 |
452. Joke Problem. Two housewives together bought a large piece of meat at the market and began to divide it into two equal parts. They have no scales, but they do have a knife and an ax. How can they divide the meat so that each of them gets, in her opinion, no less than half of the piece? | $\triangle$ One of them should divide the meat into two parts, which she considers equal, and the other should choose one of these parts. After this, both should be satisfied: the first one because she divided, and the second one because she chose. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,294 |
454. Several identical boxes together weigh 10 tons, with each of them weighing no more than 1 ton. What is the minimum number of three-ton trucks needed to haul away all this cargo in one trip? | $\triangle$ Four trucks may not be enough. For example, if there are 13 identical boxes weighing $\frac{10}{13}$ tons each, then in one of the trucks, you cannot place more than three boxes, because with four boxes, the weight is $4 \cdot \frac{10}{13}$ tons, which is more than 3 tons. In this case, with four trucks, y... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,295 |
457*. 80 students are arranged in a rectangle $8 \times 10$. All of them are of different heights. In each transverse row, the tallest student was chosen. The shortest of them turned out to be Andreev. In each longitudinal row, the shortest student was chosen. The tallest of them turned out to be Borisov. Who is taller... | $\triangle$ Consider student $B$, who stands in the same transverse row with Andreev and in the same longitudinal row with Borisov (Fig. 50). Let's denote the heights of Andreev, Borisov, and B as $h_{1}, h_{2}$, and $h_{3}$, respectively.
Since Andreev is the tallest in his transverse row, $h_{1}>h_{3}$. Since Boriso... | Andreev | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,296 |
## 4. Whose portrait am I looking at?
When I was a child, this puzzle was incredibly popular. Now it is less well-known. This puzzle has one remarkable feature: most people give the wrong answer to the question and stubbornly defend their solution despite all arguments. I remember that about 50 years ago in one compan... | 4. It is amazing how many people give the wrong answer to the question of this puzzle. They mentally put themselves in the place of the person looking at the portrait and reason as follows: “Since I have no brothers or sisters, I must be the only son of my father. Therefore, I am looking at my own portrait.”
The first... | Thepersonislookingatportraitofhisson | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,297 |
## 5.
Suppose that in the previous task, the person examining the portrait answered the question as follows: "I grew up alone, alone in the family. And yet, the son of the one in the portrait is the son of my father (you didn't mishear, it's all correct - the son!)"
Whose portrait is this person examining? | 5. In this case, the person is examining a portrait of his father. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,298 |
6. What would happen if an all-destroying cannonball hit an unbreakable pillar
Here's another riddle from my childhood that I really enjoyed. By an all-destroying cannonball, we mean a ball that sweeps away everything in its path, and by an unbreakable pillar, we mean a pillar that cannot be felled or broken. What wou... | 6. Given the conditions, the problem is logically contradictory: an all-destroying cannonball and an indestructible pillar cannot exist simultaneously. If an all-destroying cannonball existed, it would, by definition, knock down any pillar in its path. Therefore, in this case, an indestructible pillar could not exist. ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,299 |
## 7.
The following very simple problem is one of many entertaining problems that have gained widespread popularity. In a dark room stands a wardrobe, in the drawer of which there are 24 red and 24 blue socks. How many socks should be taken from the drawer to ensure that at least one pair of socks of the same color ca... | 7. Usually, the question of the problem is given an incorrect answer: 25 socks. If the problem asked how many socks should be taken from the drawer to ensure that at least 2 socks of different colors are among them, then the correct answer would indeed be such: 25 socks. But in our problem, the question is about ensuri... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,300 |
## 9.
Here is a familiar logic problem. It is known that there are more residents in New York than there are hairs on the head of any of them, and that there are no completely bald residents in New York, who do not have a single hair on their head. Does it follow that in
New York there must be at least two residents ... | 9. The answer to the first question is affirmative. Let's assume for certainty that the population of New York is 8 million people. If the number of hairs on the head of each New York resident is unique, this means that there must be 8 million different positive integers, each less than 8 million, and this is impossibl... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,302 |
## 10. Who is the Murderer?
In this story, we will talk about a caravan crossing the Sahara Desert. One day, the caravan stopped for the night. Let's denote the three main characters as A, B, and C. A hated C and decided to kill him by poisoning the water bag (the only water supply C had). Independently of A, another ... | 10. I do not think that the arguments of supporters of either opinion regarding who the murderer is can be considered "right" or "wrong." In problems of this type, it seems to me that one opinion is no better or worse than another. Personally, I believe that if anyone is to be accused of the death of the caravaner C, i... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,303 |
## 11. Another legal oddity. . . . . . . . .
Two people were on trial for murder. The jury found one of the defendants guilty and the other not guilty. The judge turned to the one who was found guilty and said, “This is the strangest case I have ever had to deal with. Although your guilt has been established beyond an... | 11. The accused were Siamese twins. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,304 |
## 13. The Clock Stopped.
Here is an excellent old puzzle. A man had no wristwatch, but at home he had accurate wall clocks, which he sometimes forgot to wind. One day, forgetting to wind the clock again, he went to visit his friend, spent the evening there, and upon returning home, he managed to
set the clock correct... | 13. When leaving home, a person winds the clock and remembers the position of the hands. Upon arriving at a friend's house and leaving, he notes the time of his arrival and departure. This allows him to find out how long he stayed at the party. Upon returning home and looking at the clock, he determines the duration of... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,306 |
## 14. The Bear Problem.
This problem has an intriguing feature: many people have heard it and know the answer, but the reasoning they use to justify it is entirely unsatisfactory. Therefore, even if you think you know the answer to the problem, check yourself by looking at the solution.
A hunter is 100 meters south ... | 14. The fur must be white, as it belongs to a polar bear inhabiting the Arctic - near the North Pole. Usually, the answer is supported by the fact that the bear in the problem's conditions must be standing at the North Pole. If you step in any direction from the North Pole, you will always be moving south. Therefore, i... | white | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,307 |
## 19. The Railway Movement Problem.
A train departs from Boston to New York. An hour later, another train departs from New York to Boston. Both trains travel at the same speed. Which one of them will be closer to Boston at the moment they meet? | 19. At the moment of meeting, the trains will be at the same distance from Boston. | Atthemomentofmeeting,thetrainswillbeatthesamedistancefromBoston | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,311 |
## 26.
So, let's start with a well-known problem. Three inhabitants of the island (A, B, and C) were talking to each other in the garden. A passerby asked A: "Are you a knight or a liar?" A answered, but so indistinctly that the passerby could not understand anything. Then the passerby asked B: "What did A say?" "A sa... | 26. Neither a knight nor a liar can say: "I am a liar" (by making such a statement, a knight would be lying, and a liar would be telling the truth). Therefore, A, whoever he was, could not have said that he was a liar. Thus, B, claiming that A called himself a liar, was definitely lying. This means B is a liar. And sin... | B | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,318 |
27.
When I first encountered the previous problem, it immediately struck me that C essentially does nothing, playing the role of a kind of "free accessory." Indeed, when B spoke, the falsity of his statement could have been established without C's involvement (see the solution to the previous problem). The following ... | 27. The answer in this problem is the same as in the previous one, but the reasoning is somewhat different.
First of all, note that B and C cannot both be knights or both be liars, since B contradicts C. Therefore, B and C cannot both be knights or both be liars: one of them is a knight, and the other is a liar. If A ... | B | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,319 |
## 28.
In this problem, there are two characters: A and B. Each of them is either a knight or a liar. A makes the following statement: "At least one of us is a liar." Who among the two characters A and B is the knight and who is the liar? | 28. Suppose A is a liar. If this were the case, then the statement "At least one of us is a liar" would be false (since liars make false statements). Therefore, in this case, A and B would both be knights. Thus, if A were a liar, he would not be a liar, which is impossible. Hence, we conclude that A is not a liar, he i... | A | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,320 |
29.
Suppose A says: "Either I am a liar, or B is a knight."
Who among the two characters, A and B, is the knight and who is the liar? | 29. This task can serve as a good introduction to logical disjunction. Let two statements $p, q$ be given. The statement "either $p$, or $q$" is true if at least one of the statements $p, q$ is true (or both). The statement "either $p$, or $q$" is false if both statements $p, q$ are false. For example, if I said on a g... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,321 |
30.
Suppose A says: "Either I am a liar, or two plus two is five." What conclusion can be drawn from this statement?
## 31 .
Before us are three islanders A, B, and C, each of whom is either a knight or a liar. Two of them (A and B) make the following statements:
A: We are all liars.
B: One of us is a knight.
Wh... | 30. The only sane conclusion one can come to is that the author of this problem is not a knight. Indeed, neither a knight nor a liar could have made the statement given in the problem. Indeed, suppose A is a knight. Then the statement "A is a liar or two plus two is five" is false, since both component statements ("A i... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,322 |
## 32.
Suppose A and B make the following statements:
A: We are all liars.
B: Exactly one of us is a liar.
Can we determine whether B is a knight or a liar?
Can we determine who C is? | 32. We cannot determine who B is, but we can prove that C is a knight.
For the same reasons as in the previous problem, A must be a liar. Therefore, at least one of the other islanders, B or C, must be a knight. Let's find out who B is. He can either be a knight or a liar. Suppose he is a knight. Then his statement th... | C | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,323 |
33.
Suppose A makes the statement: "I am a liar, and B is not a liar."
Who among the islanders A and B is a knight and who is a liar? | 33. First, note that A cannot be a knight. Indeed, if A were a knight, his statement would be true, but in it, he claims that A is a liar. Therefore, A is a liar, and his statement is false. If B were a knight, then A's statement would be true. Therefore, B is also a liar. Thus, A and B are liars. | A | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,324 |
## 34.
Once again, we have three islanders A, B, and C, each of whom is either a knight or a liar. Let's agree to call two islanders of the same type if they are both knights or both liars. Suppose A and B make the following statements:
A: B is a liar.
B: A and C are of the same type.
Who is C: a knight or a liar? | 34. Suppose A is a knight. Then his statement that B is a liar must be true, which means B must be a liar. But then B's statement that A and C are of the same type is false, so A and C are not of the same type. Therefore, C is a liar (since A is a knight). Thus, if A is a knight, then $C$ is a liar.
On the other hand,... | C | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,325 |
35.
We are again faced with three islanders A, B, and C. A makes the statement: "B and C are of the same type." Someone asks C: "Are A and B of the same type?"
What will the islander C answer? | 35. To solve this problem, we need to consider two cases separately.
First case: A is a knight. Then B and C are of the same type. If C is a knight, then B is also a knight and, therefore, of the same type as A. Therefore, C, being a truthful person, should answer "Yes." If C is a liar, then B is also a liar (since B ... | yes | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,326 |
## 36. A Minor Incident.
This puzzle is unusual. Moreover, it is based on a real incident. Once, when I was visiting the island of knights and liars, I met two locals. I asked one of them: "Is anyone of you a knight?" My question did not go unanswered, and I learned what I wanted to know.
Who was the islander I addre... | 36. Solving this problem will be helped by the information provided in the conditions of the problem after the message that the islander gave an answer to my question: my remark that after his answer I learned the true answer to my question.
Suppose the islander I was talking to (let's call him A) answered my question... | A | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,327 |
37.
Suppose you are on an island of knights and liars and come across two of its inhabitants, lazily basking in the sun. You ask one of them if their friend is a knight, and you receive an answer (yes or no). Then you ask the same question to the second islander and receive an answer (yes or no).
Should both answers... | 37. They should. If both islanders you meet are knights, they will both answer “yes”. If they are both liars, they will also both answer “yes”. If one is a knight and the other is a liar, the knight will answer “no” and the liar will also answer “no”. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,328 |
## 38. Eduard or Edwin?
This time, while strolling around the island, you accidentally come across an islander hopelessly stuck by the edge of a pond. No matter how hard you try, you can't pull him out of the mire. You remember that his name is either Edwin or Eduard, but you can't recall which. So you ask the islande... | 38. I must admit that in this task I allowed myself to play a trick on the reader. The key to the solution lies in the phrase that says that no matter how hard you tried, you couldn't "extract him from the mire." The words in quotation marks represent a pun - "extract his truths." From this, we can infer that the islan... | Edwin | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,329 |
40.
I propose to your attention an unusual problem. Two people A and B, about whom it is known that each of them is either a knight, or a liar, or a normal person, make the following statements:
A: B is a knight.
B: A is not a knight.
Prove that at least one of them is telling the truth, but it is not a knight. | 40. This task has an interesting feature. Its conditions do not allow us to determine which of the two islanders speaks the truth without being a knight: A or B. We can prove a weaker statement: at least one of the two islanders A and B speaks the truth without being a knight.
Islander A either tells the truth or does... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 36,331 |
41.
This time A and B make the following statements:
A: B is a knight.
B: A is a liar.
Prove that either one of them is telling the truth, but it is not a knight, or one of them is lying, but it is not a liar. | 41. Let's prove that if B speaks the truth, not being a knight, and if B does not speak the truth, then A lies, not being a liar.
1) Suppose that B speaks the truth. Then A is a liar and, consequently, does not speak the truth. From this, we conclude that B is not a knight. Thus, in this case, B speaks the truth, not b... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 36,332 |
## 44.
Let's first consider the married couple - Mr. and Mrs. A. They make the following statements:
Mr. A: My wife is not a normal person.
Mrs. A: My husband is not a normal person.
Who is Mr. A and who is Mrs. A - a knight, a liar, or a normal person? | 44. Mr. A cannot be a liar, because then his wife would be a knight and, consequently, could not be a normal person, which would mean that Mr. A's statement was true. For the same reason, Mrs. A cannot be a liar either. Therefore, neither Mr. A nor Mrs. A can be knights (otherwise the other spouse would be a liar). Thu... | Mr.AMrs.Anormalpeople(bothlie) | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,334 |
45.
Suppose Mr. and Mrs. A made the following statements:
Mr. A: My wife is a normal person.
Mrs. A: My husband is a normal person.
Does the answer to this problem match the answer to the previous problem? | 45. It matches. Why?
The above text has been translated into English, preserving the original text's line breaks and format. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,335 |
## 47.
One day, Alice met the Lion and the Unicorn resting under a tree. They made the following statements.
Lion. Yesterday was one of the days when I lie.
Unicorn. Yesterday was one of the days when I also lie.
From these two statements, Alice (a very smart girl) was able to deduce what day of the week it was yes... | 47. The Lion could say that he lied the day before only on Monday and Thursday. The Unicorn could say that he lied the day before only on Thursday and Sunday. Therefore, they could both claim that they lied the day before only on Thursday. | Thursday | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,337 |
48.
Another time, Alice met a Lion. He made two statements:
1) I lied yesterday.
2) After tomorrow, I will lie for two days in a row.
On what day of the week did Alice meet the Lion? | 48. From the first statement by Lev, it follows that Alice met him on Monday or Thursday. From the second statement, it follows that the meeting day was not Thursday. Therefore, the meeting took place on Monday. | Monday | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,338 |
49.
On which days of the week can Leo make the following statements:
1) I lied yesterday.
2) I will lie tomorrow. | 49. Lev cannot make such statements on any day of the week. The first statement he could only make on Monday and Thursday, the second - only on Wednesday and Sunday. Therefore, he could not have made both statements on any day of the week. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,339 |
50.
On which days of the week can Leo make the following single statement: “I lied yesterday, and I will lie tomorrow.” Warning! The answer to this problem does not match the answer to the previous problem.
## B. TRALALALA AND TRULALALA
Once, for an entire month, Leo and the Unicorn did not show up in the Forest of... | 50. The situation in this problem is quite different from the one we encountered in the previous problem. This example clearly shows the difference between two separate statements and one complex statement - their conjunction. Indeed, if any two statements $X, Y$ are given, then the truth of one complex statement «$X$ ... | MondayorWednesday | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,340 |
51.
One day, Alice met both brothers together, and they made the following statements:
F i r s t. I am Tralala.
S e c o n d. I am Trulala.
Who was actually Tralala and who was Trulala? | 51. If the first statement is true, then the first brother is called Tra-la-la. Then the second brother is called Tru-ly-ly, and the second statement is also true. If the first statement is false, then the first brother is called Tru-ly-ly, the second is Tra-la-la, and therefore the second statement is also false. Thus... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,341 |
52.
On another day of the same week, the brothers made the following statements:
First. I am Tra-la-la.
Sec ond. If that is true, then I am Tru-li-li!
Who is Tra-la-la and who is Tru-li-li? | 52. Despite the great external similarity, this task is quite different from the previous one. The second statement is deliberately true. Since the meeting took place the day after the meeting described in the previous task, it fell on a weekday. Therefore, both statements cannot be true, from which we conclude that th... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,342 |
53.
One day, Alice met both brothers and asked one of them: "Do you lie on Sundays?" That one answered: "Yes!" Then she asked the same question to the other brother. What did he answer?
54.
On another occasion, the brothers made the following statements:
First. 1) I lie on Saturdays.
2) I lie on Sundays.
Second... | 53. The first answer was deliberately false. Therefore, Alice's meeting with the two brothers took place on a weekday. But then the other brother must have given the truthful answer and said: "No". | No | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,343 |
55.
One day, Alice met one of the brothers. He stated the following: "I am lying today, and my name is Trulala."
Which of the brothers did Alice meet? | 55. The brother's statement is deliberately false (if it were true, the brother would have been lying on the day of the meeting, leading to a contradiction). Therefore, at least one of the two statements, "I am lying today," "My name is Trulala," must be false. The first statement ("I am lying today") is true, so the s... | Tralala | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,344 |
56.
Suppose the brother Alice met said: "I am lying today or I am Trulala." Could it be determined in this case which of the brothers it was? | 56. It is possible. If the brother Alice met that day was lying, then the first statement in the disjunction would be true, which would make the entire complex statement true as well, leading to a contradiction. Therefore, on the day of the meeting with Alice, the brother was telling the truth, and his statement is tru... | Trulala | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,345 |
59.
Taking a rattle with her, Alice set off to the Forest of Forgetfulness, hoping she would manage to find at least one of the brothers. To her delight, she suddenly saw both brothers sitting under a tree and grinning. Alice approached the one who was closer and sternly demanded:
- Tell me the truth! Whose rattle i... | 59. Suppose the first brother told the truth. Then the rattle belongs to Trulala. The second brother must be lying in this case (since Alice's meeting with the brothers did not take place on Sunday), so his real name is not Trulala. Therefore, he is Tralala, and Alice should give him the rattle.
Suppose now that the f... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,348 |
Alice returned the rattle to its rightful owner, but a few days later, another brother broke the rattle again. This time, the raven did not come to scare the brothers, and they began to beat each other with all their might. Alice grabbed the broken rattle and ran out of the forest.
After some time, Alice met the White... | 60. The chances are zero. Suppose the statement of the brother Alice met is true. Then the owner of the rattle should have been lying on the day of the meeting and, consequently, could not have been the brother Alice met. On the other hand, suppose the statement of the brother Alice met is false. Then the owner of the ... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,349 |
61.
A few days later, while wandering through the forest, Alice came across one of the brothers sitting under a tree. She asked him the same question and heard in response: “The true owner of the rattle speaks the truth today.”
Alice paused to think. She wanted to assess how likely it was that the brother who said t... | 61. Shaltai-Boltai correctly assessed the chances. Suppose the brother was lying. This would mean that the rattle owner does not tell the truth on the day of the meeting. On the day of the meeting, he lies and, therefore, must be the very brother whom Alice met. Now suppose that the brother Alice met is telling the tru... | \frac{13}{14} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,350 |
62.
This time, Alice met both brothers together. She asked the first one, "Are you the owner of the rattle?" - And received the answer: "Yes." Then Alice asked the second brother, "Is this your rattle?" The second one answered, and Alice gave the rattle to one of them.
To which of the brothers did Alice hand over th... | 62. The key to the solution lies in that part of the problem statement where it is clear that, upon receiving the second brother's answer, Alice knew to whom she needed to give the rattle. If the second brother had answered "yes," then one of the brothers would have been telling the truth while the other was lying, and... | Thefirstbrother | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,351 |
## 63. First Version
Alice met one of the brothers in the forest. At least he looked like either Tralala or Trulala. Alice told him the story that Shaltay-Boltay had told her, and asked: “And who are you?” To which she received a mysterious answer: “I am either Trulala or Tralala, and today is one of the days when I l... | 63. Yes, Trululu must exist. It was with him that Alice was talking.
Indeed, the brother she met claimed that both of the following statements are true:
(1) I am either Trulala or Tralala.
(2) I am lying today.
If his claim were true, then statements (1) and (2) would be true. But then statement (2) would be true, ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,352 |
66. The Fourth Version.
According to this version, Alice met on a weekday (not on Saturday and not on Sunday) two little brothers (at least by appearance, the two little men could not be distinguished from Trololo and Tralala) and asked: "Does Trulululu really exist?" They answered her as follows:
First. Trulululu ex... | 66. Whoever the second brother might be, his statement is necessarily true. (It seems that Descartes noted: “Anyone who asserts that he exists utters a true statement.” I have never met anyone who did not exist.) Since the second statement is true and the day of the meeting was not Sunday, the first statement must be f... | Trululudoesnotexist | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,355 |
67.
Portia, from Shakespeare's comedy "The Merchant of Venice," had three caskets: one of gold, one of silver, and one of lead. One of these caskets contained a portrait of Portia. A suitor was asked to choose a casket, and if he was lucky (or wise) enough to choose the casket with the portrait, he would have the rig... | 67. The statements engraved on the golden and lead caskets are opposite, so one of them must be true. Since no more than one of the three statements is true, the statement on the lid of the silver casket is false. Therefore, the portrait is actually in the silver casket.
This problem also allows for another solution. ... | The\portrait\is\in\the\golden\casket | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,356 |
68. The First Trial
During this trial, two inscriptions were engraved on the lid of each casket. Portia explained that on each lid, no more than one statement was false.
On the gold
On the silver
On the lead
1) The portrait is not in this casket
2) The portrait was painted by an artist from Venice
3) The portrait ... | 68. The lead casket can be immediately excluded from consideration, as if the portrait were in it, then both statements engraved on the lid of the lead casket would be false. Therefore, the portrait is in the gold or silver casket. The first statements engraved on the lids of the gold and silver caskets are consistent,... | The\portrait\is\in\the\lead\casket | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,357 |
70.
The fourth and final story is the most astonishing of all and serves as an illustration of a very important logical principle.
The suitor of Portia III, mentioned in the previous story, successfully overcame all three trials and became her husband, and they lived happily together for many years, having many chil... | 70. The suitor of Portia should have understood that without information about the truth or falsity of any statement or about the truth values taken by the accepted statements, the statements do not allow one to draw any conclusion, and the object (portrait or dagger) can be anywhere. What prevents me from taking any n... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,359 |
71.
Let's start with a simple case. A major theft was committed at a warehouse. The criminal (or criminals) transported the stolen goods by car. Suspicion fell on three repeat offenders A, B, and C, who were brought to Scotland Yard for interrogation. The following was established:
1) No one other than A, B, and C w... | 71. Let us first show that at least one of A, C is guilty. If B is not guilty, then it is clear that someone from A, C (or both) is guilty, since from statement (1) it follows that no one other than A, B, and C can be guilty. If B is guilty, then he must have an accomplice (since B does not know how to drive a car). Th... | A | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,360 |
73. The Case of the Indistinguishable Twins... This case is more interesting than the previous ones. A robbery has been committed in London. Three suspects - recidivists A, B, and C - have been summoned for interrogation. Suspects A and B are twins and resemble each other so much that few people can tell them apart. Th... | 73. Suppose B is not guilty. Then one of the twins must be guilty. This twin must have an accomplice, and since B could not have been the accomplice, it must have been the other twin. But this is impossible, as one of the twins was seen in Dover at the time of the crime. Therefore, B is guilty. And since B always "goes... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,361 |
## 74.
"Based on the following facts, what conclusions would you draw?" Inspector Craig asked Sergeant MacPherson:
1) If A is guilty and B is not guilty, then C is guilty.
2) C never acts alone.
3) A never goes on a job with C.
4) No one other than A, B, and C is involved in the crime, and at least one of the three i... | 74. The guilt of V is beyond any doubt. This can be proven by any of the following arguments.
Argument one. Suppose V is not guilty. Then if A were guilty, C would also be guilty by statement (1). This would mean that contrary to statement (3), A committed the crime with C. Therefore, A must be not guilty. But then, c... | V | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,362 |
## 75. The Case of the McGregor Shop Robbery...
Mr. McGregor, the owner of a shop in London, reported to Scotland Yard by telephone that he had been robbed. Three repeat offenders, A, B, and C, suspected of the robbery, were summoned for interrogation. The following facts were established:
1) Each of the three suspec... | 75. Inspector Craig brought charges against Mr. McGregor for attempting to mislead the police: there was no robbery in reality. Here is how Inspector Craig reasoned.
First step. Suppose A was guilty. Then, by statement (2), he would have had exactly one accomplice - no more, no less. Therefore, one of B, C would be gu... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,363 |
76. The Case of the Four.
This time, four suspects in a robbery were called for interrogation: A, B, C, and D. Indisputable evidence has proven that at least one of them is guilty and that no one other than A, B, C, and D was involved in the robbery. Additionally, the following facts were established:
1) A is undoubt... | 76. If B is guilty, then by statement (2) exactly two defendants are involved in the crime. If C is guilty, then by statement (3) three defendants are involved in the crime. Since neither can be true, at least one of B and C is not guilty. Defendant A is also not guilty, so there cannot be more than two guilty individu... | D | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,364 |
78.
On charges of robbery, A, B, and C are standing trial. The following has been established:
1) If A is not guilty or B is guilty, then C is guilty.
2) If A is not guilty, then C is not guilty.
Can the guilt of each of the three defendants be determined based on this information? | 78. It is possible, and very simple. According to statement (1), if A is not guilty, then C is guilty (since if A is not guilty, the disjunction "either A is not guilty or B is guilty" is true). According to statement (2), if A is not guilty, then C is not guilty. Therefore, if A is not guilty, then C is simultaneously... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,366 |
## 79.
On charges of robbery, A, B, and C were brought to court. The following has been established:
1) At least one of the three defendants is guilty.
2) If A is guilty and B is not guilty, then C is not guilty.
This information is insufficient to prove the guilt of each of the three defendants individually, but it... | 79. Two defendants, one of whom must be guilty, are V and C. Indeed, suppose A is not guilty. Then, by statement (1), V or C must be guilty. On the other hand, suppose A is guilty. If V is guilty, then at least one of V and C is definitely guilty. But suppose V is not guilty. Then A is guilty, and V is not guilty. Ther... | VorC | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,367 |
80.
This case is more interesting than the previous ones. There are four defendants: A, B, C, D. The following has been established:
1) If A and B are both guilty, then C was an accomplice.
2) If A is guilty, then at least one of the accused B, C was an accomplice.
3) If C is guilty, then D was an accomplice.
4) If ... | 80. First, let's prove that if A is guilty, then C is guilty. Suppose A is guilty. Then, by statement (2), either B or C is guilty. If B is not guilty, then C must be guilty. But suppose B is guilty. Then both A and B are guilty. Therefore, by statement (1), C is also guilty. This proves that if A is guilty, then C is ... | D | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,368 |
81.
And in this case, there were four defendants: A, B, C, and D. The following has been established:
1) If A is guilty, then B was an accomplice.
2) If B is guilty, then either C was an accomplice or A is not guilty.
3) If D is not guilty, then A is guilty and C is not guilty.
4) If D is guilty, then A is guilty.
... | 81. All defendants are guilty. Indeed, by statement (3), if D is not guilty, then A is guilty. By statement (4), if D is guilty, then A is guilty. Therefore, regardless of whether D is guilty or not, defendant A must be guilty. Then, by statement (1), B is also guilty. From statement (2), we conclude that either C is g... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,369 |
82. Wouldn't it have been better to remain silent?
On a small island, a person was being tried for a crime. The court knew that the defendant was born and raised on a neighboring island of knights and liars. (Let's recall that knights always tell the truth, while liars always lie.) The defendant was allowed to make on... | 82. Quite reasonable: it helped the defendant clear all suspicions! Indeed, let's assume the defendant is a knight. Then his statement is true, and the guilty one is a liar. Therefore, the defendant must be not guilty. On the other hand, let's assume the defendant is a liar. Then his statement is false, so the one who ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,370 |
83. The Mysterious Accuser.
In another case on the same island, two local residents, X and Y, were tried for a certain crime. The case was highly unusual because it was known that the accuser was either a knight or a liar. At the trial, the accuser made the following two statements:
1) $X$ is guilty.
2) X and Y canno... | 83. Suppose the prosecutor were a liar. Then statements (1) and (2) would be false. But if statement (1) is false, then X is not guilty, and if statement (2) is false, then both X and Y are guilty. Thus, X would have to be guilty and not guilty at the same time, which is impossible. Therefore, the prosecutor must be a ... | X | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,371 |
84.
Suppose the prosecutor from the previous problem made the following two statements in court:
1) Either $\mathrm{X}$ is guilty, or $\mathrm{Y}$ is guilty.
2) $X$ is not guilty.
What conclusion would you draw based on these statements? | 84. If the accuser were a liar, then
1) X and Y would both be guilty;
2) X would be guilty.
And in this case, we would again arrive at a contradiction. Therefore, the accuser is a knight, X is not guilty, and Y is guilty. | Y | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,372 |
85.
Suppose the prosecutor from problem 83 made the following two statements in court:
1) Either $X$ is not guilty, or $Y$ is guilty.
2) $X$ is guilty.
What conclusion would you draw based on these statements? | 85. Suppose the accuser were a liar. Then statement (1) is false, so $\mathrm{X}$ is guilty and Y is not guilty. Therefore, X is guilty. But statement (2) is also false, so $\mathrm{X}$ is not guilty, and we arrive at a contradiction. Thus, in this problem, as in the previous one, the accuser is a knight. Then, by stat... | X | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,373 |
86.
This incident took place on the island of knights, liars, and normal people. Let's recall that knights always tell the truth, liars always lie, and normal people sometimes tell the truth and sometimes lie.
Three islanders, A, B, and C, appeared in court. It is known that the crime could have been committed by on... | 86. Defendant A cannot be a knight, for if he were a knight, he would be guilty and would not lie, claiming that he is not guilty. Defendant A cannot be a liar either, for if he were a liar, his statement would be false, and he would be guilty and, consequently, a knight. Therefore, A is a normal person and not guilty.... | B | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,374 |
87.
This case, the most interesting of all, outwardly resembles the previous ones, but in reality, it is fundamentally different from them. It also took place on the island of knights, liars, and normal people.
The main characters were the defendant, the prosecutor, and the defense attorney. The trial proceeded in a ... | 87. Until Craig Arrived*. First, note that A cannot be a liar, because if he were a liar, his statement would be false, and consequently, he would be guilty. This would lead to a contradiction with the condition of the problem, which states that a liar is not guilty. Therefore, A is either a knight or a normal person.
... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,375 |
88.
You are interviewing three forest inhabitants A, B, C. It is known that exactly one of them is a werewolf. In conversation with you, they state:
A: C is the werewolf.
B: I am not the werewolf.
C: At least two of us are liars.
Our task consists of two parts.
a) Is the werewolf a knight or a liar?
b) If you h... | 88. C is either a knight or a liar. Suppose C is a knight. Then at least two of the three islanders are liars. Therefore, they must be A and B. From this, we conclude that B is a werewolf (since B claims not to be a werewolf, but B is a liar according to the proof). Thus, if C is a knight, the werewolf is a liar (since... | A | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,376 |
89.
You are interviewing three forest inhabitants A, B, and C again. It is known that each of them is either a knight or a liar, and among them, there is exactly one werewolf. In conversation with you, they state:
A: I am the werewolf.
B: I am the werewolf.
C: No more than one of us is a knight.
Conduct a full cla... | 89. Let's first prove that C is a knight. Suppose C was a liar. Then his first statement would be false, which means at least two of the three other islanders would be knights. This would mean that A and B must both be knights (since by assumption C is a liar). Consequently, their statements would be true, and they wou... | C | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,377 |
## 91.
This time A and B made the following statements:
A: At least one of us is a liar.
B: $C-$ is a knight.
It is known that exactly one of them is a werewolf and that he is a knight. Who is the werewolf? | 91. A must be a knight for the same reasons that V was a knight in the previous problem, namely: if A were a liar, then the statement that at least one of the three is a liar would be true, and we would arrive at a contradiction (the statement of a liar would be true). Since A is a knight, his statement is true, so at ... | A | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,379 |
92.
In this problem, A and B made the following statements:
$A:$ At least one of us is a liar.
B: C is a werewolf.
And in this problem, it is known that exactly one of them is a werewolf and that he is a knight. Who is the werewolf? | 92. From statement A, it follows that A must be a knight and at least one of the three must be a liar. If B were a knight, then C would be a werewolf and, therefore, another knight, but then all three would be knights. Therefore, B is a liar. But then C is not a werewolf. Since it is known that the werewolf is a knight... | A | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,380 |
93.
In this problem, it is known that out of three forest inhabitants, exactly one is a werewolf, who is a knight, and the other two are liars. Only V made a statement: "C is the werewolf."
Who is the werewolf? | 93. If V were a knight, then S would be a werewolf and a knight, which means there would be two knights. Therefore, V is a liar, and S is not a werewolf. Moreover, V, being a liar, is not a werewolf. Thus, A is the werewolf. | A | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,381 |
94.
In this task, distinguished by its elegant simplicity, there are only two characters: A and B. Only one of them is a werewolf. A and B made the following statements:
A: The werewolf is a knight.
B: The werewolf is a liar.
Whom of them would you choose as your companion?
## B. HOW TO CHOOSE OR CONQUER HEBECTY | 94. You should choose A. Suppose B is a knight. Then his statement is true. Therefore, the werewolf is a liar, so B cannot be the werewolf. Suppose B is a liar. Then his statement is false, which means the werewolf is actually a knight. Therefore, in this case B cannot be the werewolf either. | A | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,382 |
95. How to convince her?
Suppose you are one of the inhabitants of the island of knights and liars. You have fallen in love with a girl and want to marry her. But your beloved has strange tastes: for some inexplicable reason, she does not want to marry a knight and will only consent to be the wife of a liar. Moreover,... | 95. All you need to say is: "I am a poor liar." From this statement, your beloved will immediately conclude that you are not a knight (since a knight would not lie and claim to be a poor liar). Therefore, you must be a liar, and since your statement is false, you are not a poor liar. But you are a liar, so you must be ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,383 |
## 97. How to Choose a Bride?
This time you find yourself on an island of knights and liars. Every inhabitant of this island is either a knight or a liar. You fall in love with one of the beautiful islanders, a girl named Elizabeth, and want to marry her. But you would like to know who she is (naturally, you wouldn't ... | 97. This task has several solutions. The simplest one is as follows. You ask the brother of your beloved: “Are you and Elizabeth of the same type?” If he answers “yes,” then Elizabeth must be a knight regardless of whether her brother is a knight or a liar. If he answers “no,” then Elizabeth must be a liar regardless o... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,385 |
## 98. How to Choose a Bride on the Island of Bahava?
This time, you find yourself on the island of Bahava, where knights always tell the truth, liars always lie, and normal people sometimes tell the truth and sometimes lie. Let's recall that on the island of Bahava, women enjoy equal rights with men. Among the women,... | 98. This problem also admits several solutions. The simplest and most elegant of the solutions known to me consists in asking one of the sisters (say, A): "Is B ranked lower than C?"
Suppose A answers "yes." Then you choose B as your bride, reasoning as follows. Suppose A is a knight. Then B is indeed ranked lower tha... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,386 |
99.
Suppose the criminal is a liar (which is known to the court) and you are also a liar (which is unknown to the court), but you are not guilty of the crime you are accused of. You are given the right to say one and only one sentence. Your goal is to convince the jury not only that you are not a liar, but also that ... | 99. All suspicions could be removed from you by a single statement: "I am guilty." You, being a liar, could make such a statement in court, as it is false, and it would remove suspicions from you, since the jurors, well-versed in logic, would reason as follows. If you were indeed guilty, then you would be a liar (since... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,387 |
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