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101.
In this task, we will assume that the criminal is a knight. (Our assumption is internally consistent: to commit a crime, one does not necessarily have to lie.) Suppose also that you are a knight (which is unknown to the jury), but you are not guilty of the crime. What would you state in court? | 101. This problem is in a sense "dual" to problem 99 (and even somewhat easier). You need only state in court: "I am not guilty." Upon hearing your statement, the jurors would reason as follows. If you are a knight (which they do not know), then your statement is true. Consequently, you are not guilty. If, however, you... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,389 |
## 102.
This riddle is a bit more challenging. Suppose the criminal is not a normal person, but either a knight or a liar. You are not guilty. What statement, which could come from a knight, a liar, or a normal person, would you make in court to convince the jury of your innocence? | 102. One solution is for you to state in court: "Either I am a knight and not guilty, or I am a liar and guilty." Let's rephrase your statement more simply: "I am either an innocent knight or a guilty liar." Upon hearing such a statement, the jurors would reason as follows.
First step. Suppose he is a knight. Then his... | notguilty | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,390 |
## 103.
Here is a much simpler problem. It is known that the criminal is not a normal person. You are not a criminal, but you are quite normal. What statement, which could not have come from either a guilty knight or a liar, would you make in court to convince the jury of your innocence? | 103. This task is solved very simply. You need to state in court: "I am a liar." Neither a knight nor a liar could make such a statement. Therefore, you are a normal person and, consequently, not guilty. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,391 |
## 105.
This task is, in a sense, "dual" to the previous one. It is known that the criminal is not a normal person, you are not guilty, but not a knight. Suppose that for some reason you are not averse to acquiring the reputation of a liar or a normal person, but you hold knights in contempt. Could you convince the ju... | 105. You could say: "I am a guilty liar." Upon hearing your statement, the jurors would reason as follows: "Clearly, he is not a knight. Therefore, he is either a normal person or a liar. If he is a normal person, then he is not guilty. Suppose he is a liar. Then his statement is false, and he could be a guilty liar. T... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,393 |
106.
You, an inhabitant of the island of knights, liars, and normal people, are in love with the king's daughter Margozita and want to marry her. The king does not want his daughter to marry a normal person and gives her paternal advice: “Believe me, dear, you really shouldn't marry a normal person. Normal people are... | 106. No number of statements would be enough to convince the king of your insanity. Indeed, any of your statements, no matter how many there are, could have been made by a normal person, since a normal person makes both true and false statements. Therefore, you will not be able to marry the king's daughter! What a pity... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,394 |
## 107.
On another island of knights, liars, and normal people, the king held opposite views and gave his daughter different paternal advice: “My dear, I don’t want you to marry any knight or liar. I would like your husband to be a solid, reputable normal person. You should not marry a knight because all knights are h... | 107. In both cases, a single statement would suffice. The King could be convinced by the true statement “I am not a knight” (such a statement could not belong to either a knight or a liar) and the false statement “I am a liar.” | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,395 |
108.
Here is a more complex version of the previous problem. Its solution represents an alternative (though overly complex) solution to the previous problem, but to solve it, the solution to the previous problem alone is not enough.
Suppose you are a resident of the island of knights, liars, and normal people, and y... | 108. In connection with this task, I would like to note that if you make the first statement, the king will learn that although you are a normal person, you have just made a true statement. If you make the second statement, the king will learn that although you are a normal person, you have just made a false statement.... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,396 |
109.
For each of two people A and B, it is known that they are either a knight or a liar. Suppose A makes the following statement: "If I am a knight, then B is a knight."
Can it be determined who A and B are: who is the knight and who is the liar?
110
A is asked: "Are you a knight?" He replies: "If I am a knight, ... | 109. These four problems are based on the same idea, which is as follows. Let $P$ be any statement, and let A be any inhabitant of the island of knights and liars. Then if A makes the statement: "If I am a knight, then $P$," he must be a knight, and the statement $P$ must be true! This is hard to believe, and we will p... | A | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,397 |
113.
Regarding A and B, it is known that each of them is either a knight or a liar. A states: "If B is a knight, then I am a liar." Who is A and who is B? | 113. A must be a knight, and B must be a liar.
First, let's prove that only a knight can make a statement of the form "If $P$, then I am a liar." Let's recall that a true statement follows from any statement. Therefore, if the statement "I am a liar" is true, then the entire statement "If $P$, then I am a liar" is als... | A | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,398 |
114.
Two individuals X and Y are being tried for participation in a robbery, A and B are testifying in court. It is known about A and B that each of them is either a knight or a liar. During the trial, the witnesses made the following statements:
A: If X is guilty, then Y is guilty.
B: Either X is not guilty, or Y ... | 114. And in fact, A asserts: "It is not true that X is guilty and Y is not guilty." But this is the same as if A asserted: "Either X is not guilty, or Y is guilty." Therefore, A and B in fact assert the same thing, but express their thought differently. Thus, the statements given in the problem are either both true or ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,399 |
## 115.
Three inhabitants A, B, and C of the island of knights and liars were interviewed, during which they made the following statements:
A: B is a knight.
B: If A is a knight, then C is a knight.
Can it be determined who among A, B, and C is a knight and who is a liar?
B. LOVE AND LOGIC | 115. Suppose A is a knight. Then B is also a knight (according to A's statement). Therefore, B's statement "If A is a knight, then C is a knight" is true. But (by assumption) A is a knight. Therefore, C is a knight (under the assumption that A is a knight).
Thus, we have proved that if A is a knight, then C is a knigh... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,400 |
116.
Assume the following two statements are true:
104
1) I love Betty or I love Jane.
2) If I love Betty, then I love Jane.
Does it necessarily follow from them that I love Betty? Does it necessarily follow from them that I love Jane? | 116. From the statements given in the problem, it does not follow that I love Betty, but it does follow that I love Jane. The fact that I love Jane can be confirmed, for example, by the following reasoning.
I either love Betty or I do not love her. If I do not love Betty, then according to condition (1) I must love Ja... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,401 |
## 117.
Suppose someone asks me: "Is it true that if you love Betty, then you also love Jane?" I answer: "If this is true, then I love Betty."
Does it follow that I love Betty? Does it follow that I love Jane? | 117. This time, it does not follow from the conditional statement that I love Jane, but it does follow that I love Betty. Indeed, suppose that I do not love Betty. Then the statement "If I love Betty, then I love Jane" must be true (since any statement follows from a false statement). But according to the conditions of... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,402 |
118.
This time we have two girls: Eva and Margaret. I am asked: "Is it true that if you love Eva, then you also love Margaret?" I answer: "If this is true, then I love Eva, and if I love Eva, then this is true."
About which girl can we confidently say that I love her? | 118. From the conditions of the problem, it follows that I should love both Eva and Margaret. Let $P$ be the statement "If I love Eva, then I love Margaret." We know:
1) If $P$ is true, then I love Eva.
2) If I love Eva, then $P$ is true.
Solving the previous problem, we have confirmed: from (1) it follows that I love... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,403 |
## 119.
This time, we will be presented with three girls: Sue, Marcia, and Diana. Suppose the following is known.
1) I love at least one of these three girls.
2) If I love Sue but not Diana, then I also love Marcia.
3) I either love both Diana and Marcia, or I don't love either of them.
4) If I love Diana, then I als... | 119. I must love all three girls. This can be proven in different ways. Let's present one of them.
According to condition (3), I love both Diana and Marcia, or I don't love either of them. Suppose I don't love either Diana or Marcia. Then, according to condition (1), I must love Sue. So, I love Sue, but I don't love D... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,404 |
## 120.
This task, although simple, is somewhat unexpected.
Suppose I am either a knight or a liar and make the following two statements:
1) I love Linda.
2) If I love Linda, then I love Katy.
Who am I: a knight or a liar? | 120. I must be a knight. If I were a liar, then statements (1) and (2) would be false. Suppose statement (2) is false. Then I would have loved Linda, but I would not have loved Katya. So, I would have loved Linda, which means statement (1) would be true. Therefore, it is impossible for both statements (1) and (2) to be... | Imustbeknight | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,405 |
## 121. A New Version of an Old Proverb . .
An old English proverb says: "A watched pot never boils." As I have found, this statement is false. Once I watched a pot standing on a hot stove, and the pot did boil.
What if we amend the old proverb, for example, like this: "A watched pot never boils, unless it is watched... | 121. To say: “P is false if not $Q$” is the same as saying: “If $P$, then $Q$”. (For example, the statement “I won't go to the movies if you don't go with me” is equivalent to the statement “If I go to the movies, then you will go with me.”) Therefore, the “corrected” version of the proverb “The pot won't boil under wa... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,406 |
## 122. Is there a treasure on this island?
On a certain island inhabited by knights and liars, a rumor spread that treasures were buried there. You arrive on the island and ask one of the locals (let's call him A) if there is gold on his island. In response to your question, A states: "There are treasures on this isl... | 122. It is impossible to determine whether A is a knight or a liar. However, the treasures must be on the island.
To solve this and other problems in the series "Are there treasures on this island?", we will establish once and for all the following main principle: if the speaker (either a knight or a liar) states "I a... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,407 |
## 123.
In the previous task, a native of Island A of knights and liars voluntarily provided you with information. Suppose now that you asked A: "Is the statement that you are a knight equivalent to the statement that treasures are hidden on this island?" If A had answered "yes," the problem would have reduced to the ... | 123. Yes, we could: there are no treasures on the island.
Let $G$ be the statement that treasures are buried on the island, and $K$ be the statement that A is a knight. By answering your question negatively, A is essentially stating that $G$ is not equivalent to $K$. Suppose A is a knight. Then $G$ is indeed not equiv... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,408 |
## 124. How I Got Rich.
Unfortunately, the story I want to tell you is not true. But since it is interesting, I still want to tell it to you.
In the ocean (which one exactly - I don't remember) there are three islands located not far from each other: A, B, and C. I managed to find out that at least one of them has bu... | 124. We know in advance that there are no treasures on Island A, that the treasures are buried either on Island B or on Island C, and that if there is at least one normal inhabitant on Island A, then the treasures are buried on both Island B and Island C.
I asked a randomly selected islander: “Is the statement that yo... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,409 |
## 125.
Once, I happened to visit another island of knights, liars, and normal people. According to rumors, unspeakable treasures were buried on that island, and I wanted to find out the truth. The king of the island (a knight) kindly introduced me to three of his subjects, A, B, and C, and informed me that no more th... | 125. To solve this problem, it is sufficient to use the fundamental principle twice (see the explanation of it in the solution to problem 123).
One question will be enough for you to determine which of the three islanders is definitely not a normal person. Addressing A, you ask him: "Is the statement that you are a kn... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,410 |
## 126. Can you reason logically?
Suppose the population of two neighboring islands consists only of knights and liars (there are no normal people on the islands). You are told that on one island there is an even number of knights, and on the other, an odd number of knights. You are also informed that on the island wi... | 126. Without the fundamental principle at your disposal, solving this problem would be quite difficult. However, the fundamental principle allows you to easily "dispose" of the problem. I assume you are familiar with the following properties of integers: the sum of two even numbers is even, and the sum of two odd numbe... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,411 |
127.
Once I came into possession of a casket, on the lid of which was engraved the following inscription:
This casket was not made by one of Bellini's sons
Whose work is this casket: Bellini, Cellini, or one of their sons? | 127. The casket made by Bellini. Indeed, if one of Bellini's sons had made it, then the statement engraved on the lid of the casket would be false, which is impossible. If the casket had been made by either Cellini or Cellini's son, then the statement would be true, which is also impossible. Therefore, the casket was m... | Bellini | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,412 |
132.
In one museum, I had the chance to see a pair of caskets adorned with the following inscriptions:
On the gold one
Both caskets in this set were made by members of the Cellini family
On the silver one
Neither of these caskets was made by either a son of Bellini or a son of Cellini
Whose work is each of the tw... | 132. The statement engraved on the lid of the golden casket cannot be true, for otherwise we would have arrived at a contradiction. Therefore, the golden casket was made by someone from the Cellini family. Since the inscription on the golden casket is false, both caskets could not have been made by members of the Celli... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,417 |
## 137.
The story I am about to tell in this problem is truly amazing. Once I saw two caskets, a golden one and a silver one, and I wanted to find out whether at least one of them was made by Bellini. After reading the inscription on one of the caskets, I could not deduce from it that at least one of the two caskets t... | 137. This task, as well as the next three tasks, allows for multiple solutions. One of the possible solutions is to decorate the lids of the caskets with the inscription: "Either both caskets were made by Bellini, or at least one of them was made by someone from the Cellini family."
Neither the father nor the son of t... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,422 |
138.
Once, I had the opportunity to see two boxes with identical inscriptions on their lids. Knowing both inscriptions, I was able to conclude that both boxes were made by Cellini, although the inscription on each box individually did not allow me to assert that even one box was made by Cellini.
What inscription, in... | 138. One of the solutions is as follows. On the lids of the caskets, it was engraved: "At least one of these caskets was made by Cellini's son." If these statements were true, then at least one casket would have been made by Cellini's son. But this is impossible, since Cellini's son does not engrave true statements on ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,423 |
## 140.
Especially highly valued are pairs of caskets that meet the following conditions:
1) From the inscriptions on the lids, it can be concluded that one of the caskets was made by Bellini and the other by Cellini, but it is impossible to determine which casket is whose work.
2) The inscription on the lid of eithe... | 140. One possible solution is as follows.
The inscription on the lid of the golden casket: "These caskets were made by Bellini and Cellini if and only if the silver casket was made by a member of the Cellini family."
The inscription on the lid of the silver casket: "The golden casket was made by a member of the Celli... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,424 |
## 141. An Extraordinary Adventure.
In my youth, before I was married, I found myself in Florence. Browsing through a local newspaper out of boredom, I unexpectedly noticed an advertisement: "Urgently requires a logician" (fortunately, it was printed in English, as I do not speak Italian at all). Out of curiosity, I w... | 141. Box A is part of a set with box D, so if we tried to form a set from box A and box C, we would arrive at the following contradiction.
Suppose that box C was made in the same set as box A. Let the inscription on the lid of box A be true. Then the inscription on the lid of box C is false. But then the inscription o... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,425 |
## 142. The First Island
On the first island, our philosopher met two native inhabitants A and B, who stated:
A: B is a knight, and this island is called Maya.
B: A is a liar, and this island is called Maya.
Can we assert that the first island is indeed called Maya? | 142. Suppose V is a knight. Then the first island is called Maya, and moreover, A is a liar. Therefore, the statement made by A is false, and thus it is not true that V is a knight and the first island is called Maya. But by the assumption, V is a knight, so the first part of the statement is true. Hence, we conclude t... | The\first\isl\is\not\Maya | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,426 |
## 143. The Second Island.
Two native inhabitants A and B of this island stated:
A: We are both liars, and this island is called Maya.
B: What is true is true.
Can we assert that the second island is indeed called Maya? | 143. It is clear that A is a liar (the statement made by A cannot belong to a knight). Since B agrees with A, B is also a liar. Since the statement made by A is false, it is not true that: 1) A and B are both liars and that 2) the second island is called Maya. But statement (1) is true, so statement (2) must be false. ... | The\\isl\is\not\the\isl\of\Maya | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,427 |
144. The Third Island.
The native inhabitants A and B of this island stated:
A: At least one of us is a liar, and this island is called Maya.
B: Absolutely correct!
Can we assert that the third island is indeed called Maya? | 144. Since B agrees with A, A and B are either both knights or both liars. If they were both knights, it would be false that at least one of them is a liar. But then the statement made by A would be false, which is impossible, since A is a knight. Therefore, A and B are both liars. This means that the statement made by... | The\third\isl\is\not\the\isl\of\Maya | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,428 |
145. The Fourth Island
Two native inhabitants A and B of this island stated:
A: We are both liars, and this island is called Maya.
B: At least one of us is a liar, and this island is not Maya.
Can we assert that the fourth island is indeed called Maya? | 145. Islander A is a known liar, as the statement made by him cannot belong to a knight. If B is a knight, then from the statement made by him it follows that the fourth island is not the Maya Island. If, however, B is a liar, then the first part of the statement made by A is true. But the entire statement of A is fals... | not\Maya | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,429 |
## 146. The Fifth Island.
The native inhabitants A and B of this island declared:
A: We are both liars, and this island is called Maya.
B: At least one of us is a knight, and this island is not Maya.
Can it be asserted that the fifth island is indeed called Maya? | 146. As in the previous task, A must be a liar, while B can be either a knight or a liar, but in either case, the fifth island is not the Island of Maya. | not\Maya | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,430 |
## 147. The Sixth Island
Two inhabitants A and B of this island stated:
A: Either B is a knight, or this island is called Maya.
B: Either A is a liar, or this island is called Maya.
Can we assert that this island is indeed called Maya? | 147. If A were a liar, then both alternatives of the disjunction he stated would be false, which would mean that B is a liar. In turn, this would mean that both alternatives of the disjunction stated by B are false, and A must be a knight. The contradiction obtained shows that A is a knight. Therefore, his statement is... | The\sixth\isl\is\the\isl\of\Maya | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,431 |
## 151. The Third Trial.
The one sitting on the throne uttered the spell: "It is not true that I am a monkey and a knight."
Who is he?
The philosopher successfully passed all three trials of the first round and was admitted to the second round. This time, the trials were conducted over three days in another room, no... | 151. Suppose the one sitting on the throne was a liar. Then it would be true that "he is neither a monkey nor a knight at the same time." Therefore, his statement would have been true, and we would have a liar capable of making true statements. The contradiction obtained shows that the one sitting on the throne is a kn... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,434 |
152. The Fourth Trial
A: At least one of us is a monkey.
B: At least one of us is a liar.
Who are A and B? | 152. B cannot be a liar, for otherwise his statement would be true. Therefore, B is a knight, so his statement is true, and A must be a liar. Then A's statement is false, and A and B are both humans. Therefore, A is a human and a liar, and B is a human and a knight | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,435 |
154. The Sixth Trial.
A: B is a liar and a monkey. I am a human.
B: A is a knight.
Who are A and B?
Our philosopher successfully passed all three trials of the second round and was admitted to the third round, which consisted of a single, albeit complex, trial. | 154. Suppose B were a knight. Then A would also be a knight (since B claims that A is a knight), and consequently, B would have to be a liar and a monkey. The contradiction obtained shows that B is a liar. From his statement, we conclude that A is also a liar. Since the first statement made by $A$ is false, it is not t... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,437 |
155.
From the Middle Sanctuary, one can exit through four doors $\mathrm{X}, \mathrm{Y}, \mathrm{Z}$, and $\mathrm{W}$. At least one of them leads to the Inner Sanctuary. Those who exit through another door are devoured by a fire-breathing dragon.
In the Middle Sanctuary, during the trial, there are eight priests A,... | 155. First, let us prove that $G$ is a knight. For this, it is sufficient to prove that his statement is true, that is, if $\mathrm{C}$ is a knight, then $\mathrm{F}$ is also a knight. We will prove this by deriving the conclusion "F is also a knight" from the premise "C is a knight."
So, suppose that C is a knight. T... | X | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,438 |
156. In the Inner Sanctuary!
Our philosopher managed to choose the right door and found himself in the Inner Sanctuary. There, on two thrones studded with diamonds, sat the two greatest priests (there were no greater priests in the entire world!). Perhaps one of them knew the answer to the Question of Questions: "Why ... | 156. The first priest cannot be a knight; he must be a liar. Since his statement is false, it is not true that he is a liar and does not know the answer to the Question of Questions. But he is a liar, so the first part of the conjunction he stated is true. Therefore, the second part of the conjunction must be false, wh... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,439 |
## 158.
Once I met a native of the island and asked him: "Does 'bal' mean 'yes' in our language?" He replied: "bal".
a) Can we conclude from our conversation what the word 'bal' means?
b) Can we conclude from our conversation whether my interlocutor is a zombie or a human? | 158. From our conversation, it is impossible to conclude what "bal" means, but we can confidently say that my interlocutor must be a human.
Suppose "bal" means "yes." Then "bal" is a truthful answer to the question of whether "bal" means "yes" in our language. Therefore, in this case, my interlocutor must be a human.
... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,441 |
159.
Imagine that while walking on an island, you meet one of the natives. Can you find out what the word "bal" means with just one question? (Remember that your interlocutor will answer all your questions either with "bal" or "da," and the word "da" in the local dialect, by a mere coincidence, sounds like the affirm... | 159. It is enough to ask the native if he is a human. Since all the indigenous inhabitants of the island consider it an honor to be called people, the native you encounter, whether a human or a zombie, will answer affirmatively to your question. If he answers “bal,” then “bal” means “yes.” If he answers “yes,” then “ye... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,442 |
160.
Suppose you are not interested in what exactly the word "bal" corresponds to (a negative or affirmative answer), but you want to know who you are talking to: a zombie or a human. Can you find this out by asking the interlocutor only one question? | 160. You can simply ask the first person you meet the same question as in problem 158, that is, ask: "Does 'bal' mean 'yes' in our language?" If 'bal' does indeed mean 'yes', then the correct answer to your question should be 'bal'. Therefore, a person will answer 'bal', while a zombie will say 'yes'. If 'bal' does not... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,443 |
161. How to make a wizard say "bal".
You are on the same island as in the previous tasks and want to marry the daughter of the local tribe's chieftain. The chieftain intends to give his daughter in marriage only to a person marked by the seal of wisdom, so you must pass a test consisting of the following.
You must as... | 161. To make the wizard answer "bal," there are several ways. For example, you can ask the wizard whether "bal" is the correct answer to the question of whether the wizard is a human. You can prove that the wizard has no other choice but to answer "bal." To simplify the subsequent reasoning, let's denote the question "... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,444 |
162.
This task is a bit more challenging than the previous ones. A rumor has spread that a treasure is buried on the island. You arrive on the island and, before starting your search, you want to find out whether anyone has actually buried a treasure or not. All the natives are well-informed about whether a treasure ... | 162. It can be done, and not just one, but in many ways. For example, you can ask the first person you meet: "If someone asks you whether there is a treasure on the island, would you answer 'bal'? As will be shown, if there is a treasure on the island, the native will answer 'bal'. If there is no treasure on the island... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,445 |
## 163. The Court.
In the language of the tribe inhabiting the island located not far from the island of zombies, the words "bal" and "da" mean "yes" and "no," but words that sound the same do not necessarily have the same meaning. Some of the island's inhabitants answer questions with "bal" and "da," while others, br... | 163. First, I will prove that witness C cannot be a zombie. Suppose that C is a zombie. Then A and B must be brothers. Therefore, they are either both zombies or both humans. Suppose A and B are humans. Then "bal" means "yes," so A gave an affirmative answer to the question of whether the defendant is guilty. Therefore... | notguilty | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,446 |
## 165. Half-Zombie.
After the trial, Inspector Craig visited a neighboring island. One part of the native population of this island consisted of ordinary people, another of zombies, and the third of so-called half-zombies. The latter included those islanders who had been cursed by sorcerers, but only partially: half-... | 165. The only one of the four possible answers ("bal", "yes", "yes", and "no") that neither a human nor a zombie could give is "no." Indeed, if the native were either a human or a zombie and answered Inspector Craig in English, he would have to say "yes." If he answered in his native language, then if "bal" means "yes,... | no | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,448 |
166. What was the answer?
On another occasion on the same island, Inspector Craig met another native and asked him: "If someone were to ask you whether it is true that two plus two is four, and you decided to answer in your native language, would you answer 'bal'?"
And this time, too, Inspector Craig did not record w... | 166. And in this task, just like in the previous one, the native must be a half-zombie, and the only answer by which Craig could determine who his interlocutor is must be "yes" (in the native dialect). If the native had answered in English, Craig would not have been able to determine who he was talking to, since both a... | yes | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,449 |
## 167.
Once I met a Transylvanian who stated: “Either I am a human, or I am sane.”
Who was he in reality? | 167. The statement made by the Transylvanian is either true or false. Suppose it is false. Then the Transylvanian is neither a human nor of sound mind. Therefore, he must be a vampire who has lost his mind. But such vampires only make true statements, leading to a contradiction. Therefore, the statement made by the Tra... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,450 |
## 170.
Once I asked a passerby: "Are you a vampire who has lost your mind?" He answered either "yes" or "no," and I found out what type he was.
Who was he? | 170. A sane person would answer my question negatively, while Transylvanians of any of the other three types would answer affirmatively. Receiving an affirmative answer to my question, I would not be able to determine which of the four types of Transylvanians my interlocutor belongs to. But according to the conditions ... | saneperson | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,453 |
171.
Once I met a Transylvanian who claimed: "I am a vampire."
Can we determine based on this statement whether he was a human or a vampire? Can we determine from the same statement whether he was in his right mind or had lost his sanity? | 171. From the given phrase, it cannot be concluded whether the person encountered was a human or a vampire, but it can be concluded that he had lost his mind. A person in their right mind could not have said about themselves that they were a vampire, and a vampire in their right mind would know that they were a vampire... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,454 |
172.
A certain Transylvanian declares: “I have lost my mind.”
a) Can it be determined from this statement whether he is in his right mind?
b) Can it be determined from the same statement whether he is a human or a vampire? | 172. The only conclusion that can be drawn from the Transylvanian's statement is that he is a vampire. A person in their right mind would not say of themselves that they have lost their mind. A person who has lost their mind would believe they are in their right mind, and being a person, could not state about themselve... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,455 |
## 173. A Clever Problem
The statement "if $Q$, then $P$" is called the converse of the statement "if $P$, then $Q$". There exist two statements $X$ and $Y$ such that each is the converse of the other, and:
1) $X$ does not follow from $Y$, and $Y$ does not follow from $X$;
2) if any resident of Transylvania states ei... | 173. I think there are quite a few such statements $X, Y$, at least not just one pair. I meant the following statements:
$X$: If I am of sound mind, then I am a human.
$Y$: If I am a human, then I am of sound mind.
Suppose a Transylvanian makes the statement $X$. Let's prove that $Y$ must be true, that is, if our Tr... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,456 |
## 174.
Let $X$ be any statement and a certain Transylvanian believes that he believes $X$ to be true. Does it follow that $X$ must be true? Suppose that our Transylvanian does not believe that he believes $X$ to be true. Does it follow that $X$ must be false? | 174. The answers to both questions of the problem are affirmative. Suppose a Transylvanian considers some statement $X$ to be true. From this, it is not hard to understand that it does not follow that $X$ must be true, as the Transylvanian may have lost his mind. But if he considers that $X$ is true, then $X$ must be t... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,457 |
175.
Suppose a Transylvanian claims: "I consider $X$ to be true." Does it follow that $X$ must be true if our Transylvanian is a human? Does it follow that $X$ must be false if our Transylvanian is a vampire?
Solving this problem establishes an important general principle! | 175. The answers to both questions of the problem (as follows from the solution of the previous problem) should be affirmative.
Suppose that according to A's statement, he considers the statement $X$ to be true. Then A indeed thinks exactly as he says. Therefore, A believes that he considers the statement $X$ to be tr... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,458 |
## 176.
Once I met two Transylvanians A and B. I asked A: "Is B a human?" A replied: "I think so." Then I asked B: "What do you think, is A a human?" What did B answer? (It is assumed that B answered either "yes" or "no".) | 176. A claims to consider B a human. B either claims to consider A a human or claims to consider A not a human. The second alternative must be excluded, as it leads to the following contradiction.
Consider the two statements.
1) A claims to consider B a human.
2) B claims to consider A not a human.
Suppose A is a hu... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,459 |
178.
Suppose you ask a Transylvanian another question: "Do you think you are reliable?" He answers either "yes" or "no." Can you conclude from the answer who your interlocutor is: a vampire or a human? Can you determine whether he is of sound mind?
## B. IS COUNT DRACULA ALIVE? | 178. The case described in this problem differs from the case considered in the previous problem. From your interlocutor's answer, it is impossible to conclude whether they are a human or a vampire, but it is possible to determine whether they are of sound mind. If the Transylvanian you encounter is of sound mind, they... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,461 |
179.
Let me remind you that, setting out for Transylvania, I primarily wanted to find out whether Count Dracula Dacian was alive. With this question, I approached the first Transylvanian I met. He replied: “If I am a human, then Count Dracula is alive.”
Can we determine from this answer whether Count Dracula is aliv... | 179. No, you can't. It's not impossible that your Transylvanian is a sane person and Count Dracula is alive. It's also possible that your interlocutor is a vampire who has lost his mind, and Count Dracula is not alive. (In reality, if you asked a vampire who has lost his mind, Dracula could be either alive or dead.) | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,462 |
180.
Another Transylvanian, in response to the same question, stated: "If I am in my right mind, then Count Dracula is alive."
Can we conclude from this answer whether Count Dracula is alive?
181
Another Transylvanian, whom I asked whether Dracula is alive, answered: "If I am a human and in my right mind, then Cou... | 180. No, it cannot. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,463 |
182.
Suppose that in response to my question, a Transylvanian declared: “If I am either a sane human or an insane vampire, then Count Dracula is alive.”
Can we conclude from this answer whether Count Dracula is alive? | 182. It follows from the answer you received that Dracula is alive.
Using the terminology of problem 177, we can formulate the Transylvanian's statement as follows: "If I am reliable, then Dracula is alive."
In Chapter 8 (see solutions to problems 109-112), we proved that a native of the island of knights and knaves,... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,464 |
184.
Is there such a statement that a Transylvanian could use to convince you, without resorting to other arguments, that Count Dracula is alive and that it is impossible to say whether the statement itself is true or false? | 184. There is such a statement: "I am reliable if and only if Dracula is alive."
In solving problem 122 from chapter 8, we proved that if a native of the island of knights and liars makes the statement "I am a knight if and only if such and such," then this "such and such" must be true (although we can say nothing abo... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,466 |
## 185.
Suppose a Transylvanian has made the following two statements:
1) I am of sound mind.
2) I believe that Count Dracula is not alive.
Can we conclude from these statements whether Count Dracula is alive? | 185. It can be. From statements (1) and (2), it would follow that Dracula is not alive.
From statement (1), it can be concluded that our Transylvanian is a human. Indeed, a vampire in a sound mind would know that he is in a sound mind and would state: "I have lost my mind." A vampire who has lost his mind would think ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,467 |
186.
Suppose a Transylvanian has made the following two statements:
1) I am a human.
2) If I am a human, then Count Dracula is alive.
Can we conclude from these statements whether Count Dracula is alive?
## V. WHAT QUESTION TO ASK? | 186. From the first statement ("I am a human"), it does not follow that the Transylvanian is a human, but it does follow that he must be of sound mind. (A human who has lost their reason would not know that they are a human. A vampire who has lost their reason would believe themselves to be a human and, lying, would sa... | CountDraculaisalive | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,468 |
187.
Can you, by asking the first Transylvanian you meet only one question, find out whether he is a vampire or a human? | 187. It is enough to ask a Transylvanian whether they are in their right mind. A human (regardless of whether they are in their right mind or have lost their reason) will answer affirmatively, while a vampire will answer negatively. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,469 |
188.
Can you, by asking the first Transylvanian you meet only one question, determine whether he is sane? | 188. Just ask the first person you meet whether they are a human, and everything will become clear. A Transylvanian in their right mind (be it a human or a vampire) will answer affirmatively, while a Transylvanian who has lost their mind will answer negatively.
In the following few problems, I will only provide the an... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,470 |
189.
What question should you ask a Transylvanian to make them answer "yes" regardless of which of the four types they belong to? | 189. One of the questions to which all Transylvanians will be forced to answer affirmatively is: "Do you consider yourself a human being?" The matter is not that all Transylvanians really consider themselves humans (only people in their right mind and vampires who have lost their sanity think so), but nevertheless, all... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,471 |
## 191.
By asking just one question (to which the answer should be either “ball” or “yes”), I could have found out from any guest whether they were a vampire or not. | 191. It is enough to ask the guest: “Is it correct to answer ‘ball’ to the question of whether you are in your right mind?” If the guest answers “ball,” then he is a human. If the guest answers “yes,” then he is a vampire. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,473 |
## 195.
By asking just one question, I could have found out whether Dracula was alive!
What is this question?
## D. DRACULA'S RIDDLE
And now we come to the climax of the entire story! The next day, I already had all the necessary information: Count Dracula was alive and in excellent health, and he was indeed the ow... | 195. Any of the following questions will allow you to find out if Count Dracula is alive.
1) Do you think that "ball" is the correct answer to the question of whether the statement that you are a human is equivalent to the statement "Dracula is alive"?
2) Is it correct to answer "ball" to the question of whether the st... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,477 |
196.
Dracula looked at me inquisitively:
- I must admit that the questions you asked my guests were quite witty. Are you surprised? Don't be! I am perfectly well informed about everything that happens in my castle. Your questions, I repeat, were very witty, though not as good as you think. Judge for yourself. To get... | 196. The Unified Principle. Let's agree to call a representative of the Transylvanian elite an aristocrat of type 1 if, in response to the question "two plus two is four?" he answers "bal." Naturally, to any other question with the correct answer "yes," a Transylvanian aristocrat of type 1 will answer "bal."
Let's agr... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,478 |
## 2. The Second Story
- Now we have jam again,-the King said to the Queen,-and you will finally be able to bake pretzels.
- How can I bake pretzels when I have no flour? - the Queen asked.
- Don't tell me the flour was stolen too?! - cried the King.
- That's exactly it! - said the Queen. - Find the one who did it and... | 2. The second story. Suppose the flour was stolen by the March Hare. Since the one who stole the flour gave truthful testimony, the March Hare said at the trial that the flour was stolen by the Bolvanchik. But we firmly know that the flour was stolen by only one of the three inhabitants of the house. Therefore, the Mar... | Sonya | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,481 |
## 3. The Third Story
- Well, here's the flour! - the King said happily. - Now you can finally bake some pretzels.
- Bake pretzels without pepper? - the Queen asked.
- Pepper? - the King asked, disbelievingly. - Are you saying that you put pepper in pretzels?
- Just a tiny bit, - the Queen replied.
- Are you saying th... | 3. The third story. If the cook had stolen the pepper, she would have known it for certain. Consequently, when giving testimony in court (when she stated that she knew who had stolen the pepper), she would have been telling the truth. Meanwhile, we firmly know that those who steal pepper never tell the truth. Therefore... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,482 |
## 4. Who Stole the Pepper?
After the Duchess's cook's innocence was proven, the King's suspicions fell on the March Hare, the Mad Hatter, and Dormouse. Soldiers were sent to their cottage, but they found no pepper during the search. Since the three suspects could have hidden the pepper somewhere, they were arrested a... | 4. Who stole the pepper? If the March Hare stole the pepper, then he was lying (because those who steal pepper always lie). Consequently, his statement about Bolvanchik is false. So, Bolvanchik also stole the pepper. But from the conditions of the problem, we know that only one person stole the pepper. Therefore, the M... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,483 |
5. So who, after all, stole the pepper?
- What can I say, it's truly a complicated case! - said the King.
This time, his suspicions, strangely enough, fell on the Griffin, Quasi the Turtle, and Omar. At the trial, the Griffin stated that Quasi the Turtle was innocent, and Quasi the Turtle claimed that Omar was guilty... | 5. So who exactly stole the pepper? Suppose Griffin was guilty. This would mean that he lied in court. Consequently, Quasi Turtle is not innocent (as Griffin claimed), but guilty. But then there would be two guilty parties, although the pepper (as mentioned in the previous problem) was stolen by only one person. Theref... | Omar | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,484 |
## 7. The Fourth Story
- What a hassle with the search for this cursed pepper! - the King said angrily. - You'd think it's impossible to bake pretzels without it! She needs pepper, you see!
- Why don't you put blotting paper in the dough for the pretzels? - he added sarcastically.
- I do, - the Queen replied, - just a... | 7. The fourth story. Suppose the Duchess stole the sugar. Then, testifying in court, she was lying. Therefore, her statement that the cook did not steal the sugar is false. In other words, the cook must have also stolen the sugar. But as we know for sure, the sugar was stolen by only one of the two accused. Therefore, ... | Thecookstolethesugar | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,486 |
## 8. The Fifth Story
- Here is your sugar,- said the King.- You can bake me some pretzels.
- How, without salt? - asked the Queen.
Indeed! The salt was stolen too! The investigation established that the theft could have been committed by the Caterpillar, Lizard Bill, or the Cheshire Cat. (Someone sneaked into the ki... | 8. The fifth story. If the Cheshire Cat ate the salt, then all three accused are lying, which contradicts the conditions of the problem. If the salt was eaten by Lizard Bill, then all three always tell the truth, which also contradicts the conditions of the problem. Therefore, the Caterpillar ate the salt (so the first... | TheCaterpillaratethesalt | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,487 |
## 9. The Sixth Story
- Here's some more salt,- said the King. Can you bake pretzels now?
- I can't,- answered the Queen. - Someone has stolen my skillet.
- Skillet! - the King cried in anger. - Well, let's find it then!
This time, the suspects were the Frog, the Groom-Pike, and the Jack of Hearts. At the trial, they... | 9. The sixth story. If the pan was stolen by Froggy, then both he and the Jack of Hearts lied, which is excluded by the conditions of the problem. If the pan was stolen by the Lackey-Perch, then both he and the Jack of Hearts lied, which is also excluded by the conditions of the problem. Therefore, the pan was stolen b... | TheJackofHearts | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,488 |
10. The Seventh Story
- Here's a frying pan, - said the King. Will you finally bake me some pretzels or not?
- Without a recipe? - asked the Queen.
- Use your usual recipe, shouted the King impatiently. - Last time your pretzels turned out wonderfully!
. If the cook had stolen the cookbook, then all three accused would have be... | TheDuchess | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,489 |
## 11. Continuation of the Seventh Story
Soon after the cookbook was returned to the Queen, it was stolen again. And once again, suspicion fell on the Duchess, the Cook, and the Cheshire Cat.
At the trial, all three gave the same testimonies as before. But this time, the one who stole the cookbook lied, while the oth... | 11. Continuation of the seventh story. The Cheshire Cat could not steal the cookbook for the same reason as in the previous problem. Suppose the Duchess stole the cookbook. Then the Cheshire Cat is lying, and the Cook is telling the truth, which contradicts the condition of the problem (if the cookbook was stolen by th... | TheCook | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,490 |
## Story Eighth
- Here is your cookbook again,-said
the King.-Now you have the recipe too. Bake me some pretzels!
- Without milk, butter, and eggs?
- Woe is me!-cried the King.-This is too much!
- But now I know for sure who stole my supplies! It was the March Hare, the Mad Hatter, and the Dormouse, shouted the Quee... | 12. The eighth story. First of all, note that Sonya could not have stolen the butter (the one who stole the butter speaks the truth, but Sonya testified in court that she stole the milk). Therefore, Sonya did not steal the milk. This means that the butter was stolen either by the March Hare or by the Bolvanchik. If the... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,491 |
## 16. The Fish-Footman and the Frog-Footman
- All this is very interesting, - said Alice, only the two cases you told me are quite different.
- Of course, my dear! And the moral of the whole thing is: to be or not to be is not the same as to be and not to be.
Alice tried to understand what the Duchess meant, but the... | 16. The Butler-Perch and the Tadpole. The information provided in the problem does not allow us to determine whether the Butler-Perch is in his right mind or not, but we will prove that the Tadpole must be in his right mind. We will reason as follows.
There are two possibilities: either the Butler-Perch is in his righ... | TheFrog-Footmanisinhisrightmind | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,495 |
18. The March Hare, the Hatter, and the Dormouse
- I've been thinking a lot about the March Hare, the Hatter, and the Dormouse,- said Alice.- The Hatter is called the Mad Hatter, but is he really mad? And what about the March Hare and the Dormouse?
- You see, my dear,- replied the Duchess, - the Hatter once mentioned ... | 18. The March Hare, the Mad Hatter, and Sonya. Suppose the Mad Hatter is in his right mind. Then he judges everything soundly. This means that the March Hare does not think that all three participants of the mad tea party are in their right minds. Therefore, the March Hare must be in his right mind because if he were n... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,497 |
## 19. The Griffin, the Mock Turtle, and the Oyster
- As you know, - began the Duchess, - we have a Griffin, a Mock Turtle, and an Oyster here.
- I didn't know you had a real oyster here, - remarked Alice. - To be honest, I only knew about the Oyster from the poem "This is the voice of the Oyster. Do you hear the cry?... | 19. The Griffin, Quasi the Turtle, and Omar. First of all, the Griffin and Quasi the Turtle must be "the same," that is, either both not in their right minds or both in a sound state of mind, since Quasi the Turtle believes that the Griffin is in his right mind. If Quasi the Turtle is in a sound state of mind, this mea... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,498 |
20. The King and Queen of Hearts
- You know,- Alice began in a barely audible whisper, looking around to make sure the Queen of Hearts wasn't nearby,- I would very much like to know if the King and Queen of Hearts are in their right minds. Do you happen to know?
- Of course!- exclaimed the Duchess.- It's a very intere... | 20. The King and Queen of Hearts. The Queen of Spades thinks that the King of Spades thinks that she is not in her right mind. If she is of sound mind, then the King indeed thinks that she is not in her right mind, which means the King must be out of his mind. If the Queen is not in her right mind, then the King actual... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,499 |
## 24. The Knave of Hearts
Alice had solved the previous problem.
- I think I now know why half the people here in the neighborhood are not in their right minds.
- Why? - asked the Duchess.
- Because they went mad trying to solve problems like the one you gave me. It's incredibly complicated!
- Is it a complicated pr... | 24. Jack of Hearts. Let's prove that if Seven is not in his right mind, then Six must be sane, and therefore, Jack of Hearts must have reasoned correctly, thinking that Six and Seven cannot both be insane.
Suppose Seven is not in his right mind. Then what Seven thinks about Five is false, so Five is sane. Consequently... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,503 |
## 26. How many gingerbread cookies did each have?
- This is what I call a diligent student, the Gryphon rejoiced. - Of course, I have another problem for you. The principle of solving it is a bit different, but I'm sure you'll manage.
This time, all three participated in the tea party: the Mock Turtle, the March Har... | 26. How many pretzels does each have? Let's call one portion all the pretzels that Sone got, no matter how many there are. Then Sone got 1 portion. The March Hare got twice as many pretzels as Sone (because Bovanishchik seated Sone in such a place where there were half as many pretzels as at the March Hare's place), so... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,505 | |
## 27. Retribution
- The Mock Turtle must be a terrible schemer, observed Alice. - He always tries to arrange things so that he gets more!
- Usually he does act that way, agreed the Griffin, - but once the March Hare and Dormouse got their revenge! That time, the Mock Turtle, as usual, was setting the table and put al... | 27. Retribution. After the March Hare ate $5 / 16$ of the gingerbread cookies, there were "1/16" left on the plate. Sonya ate $7 / 11$ of the remaining cookies, which is $7 / 11$ of $11 / 16$. Since $7 / 11 \cdot 11 / 16 = 7 / 16$, Sonya ate $7 / 16$ of all the cookies. Together with the March Hare, who ate ${ }^{5} / ... | 1014 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,506 |
## 28. How Many Favorites?
- "Here's a different kind of puzzle," said the Gryphon. "Once, the Queen of Hearts held a reception for thirty guests. She needed to distribute one hundred gingerbread cookies among the guests. Instead of cutting the cookies into pieces, the Queen preferred to give four cookies to each of h... | 28. How many favorites? This problem, usually solved using algebra, is very simple if approached in the following way. First, let's distribute 3 pretzels to each of the 30 guests of the Queen. We will have 10 pretzels left. At this point, all non-favorites will have received all the pretzels they are entitled to, while... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,507 |
## 29. Pretzels and Pretzelkins
- Here's another problem,—began the Gryphon. One day, the Mock Turtle went to the store to buy pretzels for the next tea party.
- How much are your pretzels?—he asked the store owner.
- The price depends on the size: I can offer you small pretzelkins and large pretzels. One pretzel cost... | 29. Pretzels and pretzelkins. Since each pretzel costs as much as one pretzelkin, 7 pretzels cost as much as 21 pretzelkins, and 7 pretzels and 4 pretzelkins cost as much as 25 pretzelkins. On the other hand, 4 pretzels and 7 pretzelkins cost as much as 19 pretzelkins (since 4 pretzels cost as much as 12 pretzelkins). ... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,508 |
30. At the Duchess's, the Cook's, and the Cheshire Cat's
- I want to offer you a most intriguing task,- said the Gryphon. - One day, the Mock Turtle, the March Hare, and the Dormouse decided to visit the Duchess, the Cook, and the Cheshire Cat. Upon arriving, they found that no one was home. On the kitchen table, they... | 30. At the Duchess's, the Cook's, and the Cheshire Cat's. The Cheshire Cat must find 2 pretzels on the tray: after eating half of the pretzels and one more pretzel, nothing will be left on the tray. Sonya must find 6 pretzels on the tray: after eating half of the pretzels and one more pretzel, 2 pretzels will be left f... | 30 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,509 |
31. How many days did the gardener work?
- I want to give you a task, - said the Griffin. - Usually it is solved using algebra, but if you use my method, you will do wonderfully without it!
Once the King hired one of the spade suit gardeners for twenty-six days to do some work in the garden. The King set the conditio... | 31. How many days did the gardener work? Working diligently, the gardener can earn a maximum of $3 \cdot 26=78$ pretzels. He earned only 62 pretzels. Therefore, he did not receive 16 pretzels because he was slacking off. Each day the gardener slacked off, he lost 4 pretzels (the difference between the 3 pretzels he cou... | 22 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,510 |
33. How many people got lost in the mountains?
Alice and the Griffin had to wait several minutes before the Tortoise Quasi gathered his strength and could continue.
- You see,一began the Tortoise Quasi.
- I don't see anything!- the Griffin cut in.
The Tortoise Quasi did not respond, only grabbing his head with his fr... | 33. How many people got lost in the mountains? Let's call one portion the amount of supplies one person consumes in a day. Initially, 9 people had 45 portions (a 5-day food supply). On the second day, they had only 36 portions left. On the same day, they met a second group, and the 36 remaining portions were enough for... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,512 |
36. How long will it take to get out of the well?
- As you wish, - agreed Quasi-Tortoise. I'll tell you a problem about a frog that fell into a well.
- But that's a very old problem! - objected the Griffin. - It has a very long beard! Don't you know any new problems?
- I haven't heard this problem, - Alice spoke up fo... | 36. How long does it take to get out of the well? Those who think the frog will get out of the well in 30 days are wrong: the frog could have gotten out of the well by evening on the 28th day. Indeed, in the morning of the 2nd day, the frog is 1 foot above the bottom of the well, in the morning of the 3rd day - 2 feet,... | 28 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,515 |
37. Will the cyclist make it to the train?
- Wasn't the previous problem sad? the Turtle Quasi asked. - Just think! The poor frog spent so many days in a dark well! And to get out of there, she had to undertake a climb like a real mountaineer!
- Nonsense! - the Griffin interrupted him. - The saddest part of the whole ... | 37. Will the cyclist make it to the train? The cyclist reasoned incorrectly: he averaged distances, not time. If he had traveled at 4 miles per hour, 8 miles per hour, and 12 miles per hour for the same amount of time, his average speed would indeed have been 8 miles per hour. However, he spent more time climbing the h... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,516 |
38. Did the passenger miss the train?
- Poor thing! - Quasiturtle sobbed. Think of it! If he were a little smarter, he could have left earlier and caught the train!
- I remembered the cyclist, - Quasiturtle continued. - The train left the station eleven minutes late and traveled at a speed of ten miles per hour to the... | 38. Did the passenger miss the train? The passenger arrived at the first station one minute after the train had left. Ten miles per hour is one mile in 6 minutes or one and a half miles in 9 minutes. Thus, the train arrived at the next station 8 minutes after the passenger arrived at the first station. The train stoppe... | Yes | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,517 |
39. How far is it to school?
All the time while Alice and the Griffin were solving the previous problem, the Tortoise Quasi was weeping inconsolably.
- Can you tell me what is sad about this problem? - the Griffin barked at him angrily. - After all, the passenger caught up with the train. Or did I misunderstand somet... | 39. How far is it to school? The difference in time between being 5 minutes late and arriving 10 minutes before the start of the lesson is 15 minutes. Therefore, if the boy walks to school at a speed of 5 miles per hour, he will save 15 minutes (compared to how long it would take him to walk if he went at a speed of 4 ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,518 |
## 42. The Appearance of the First Spy
- As I have already told you, - began the King, - in that distant country, knights always spoke the truth and never lied, while liars always lied and never told the truth. One day, the entire population of the country was in extraordinary excitement: it became known that a spy fr... | 42. The appearance of the first spy. $C$ cannot be a knight by definition, as no knight would lie and claim to be a spy. Therefore, $C$ is either a liar or a spy. Suppose $C$ is a spy. Then $A$'s statement is false, so $A$ is a spy (but $A$ cannot be a spy since $C$ is the spy), and the only one who can be a knight is ... | A | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,521 |
## 43. The Stupid Spy
- The previous task wasn't too difficult, - said Alice, solving the first problem.
- The tasks will gradually become more complex, promised the King. - The book I read in childhood was beautifully written and followed the principle of "from simple to complex." The next two tasks will also seem ea... | 43. The Stupid Spy. A false statement incriminating the spy could be, for example: “I am a liar.”
A Knight never lies and therefore would not claim to be a liar. On the other hand, a Liar never tells the truth and would not confess to being a liar. Only a Spy can make a false confession, claiming to be a liar. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,522 |
## 44. Another Foolish Spy
- The foolish spy was exposed and sent to the dungeon, - said the King, - but soon another spy infiltrated the country. He was also caught, but the officers who arrested him were not sure that he was a spy. This time, the spy gave true testimony during the interrogation, but did so in such a... | 44. Another foolish spy. A true statement exposing the spy could have been, for example: “I am not a knight.” Indeed, neither a knight nor a liar could say such a thing. A knight never lies and would not claim to be not a knight. A liar always lies and would not admit to being not a knight. Therefore, only a spy could ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,523 |
## 45. The Cunning Spy
- The next spy who infiltrated the country, said the King, turned out to be much smarter than his predecessors. He was also arrested along with two native residents, one of whom was a knight, and the other a liar. All three appeared in court. The court knew that one of the accused was a knight, ... | 45. The Crafty Spy. If A had answered the judge's question with "yes," he would have incriminated himself as the spy, because the judge (along with the jury) could reason as follows:
"Suppose B is the spy. Then all three accused would have given truthful testimonies, which is impossible since one of them is a liar. Th... | no | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,524 |
## 46. Who is Murdoch?
- A spy named Murdoch has infiltrated the country. He was arrested along with two indigenous residents (we will denote the defendants with Latin letters $A, B$, and $C$). One of the indigenous residents, mistakenly arrested along with the spy, was a knight, and the other was a liar. Of the three... | 46. Who is Murdoch? Since $A$ claims that he is a spy, $A$ is either a liar or a spy. Similarly, since $C$ claims that he is a spy, $C$ is either a liar or a spy. Therefore, out of the two defendants $A$ and $C$, one is a liar and the other is a spy. Thus, $B$ is a knight and gave truthful testimony in court: $A$ is th... | A | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,525 |
## 47. The Return of Murdoch - Murdoch was thrown into the dungeon, - continued
the King, - but soon escaped and left the country, but then he infiltrated the country again, thoroughly disguised so that no one could recognize him. And this time he was arrested along with one knight and one liar. All three (let's denot... | 47. The Return of Murdoch. If $A$ is Murdoch, then all three statements are true, which is impossible, as one of the three defendants is a liar. If $C$ is Murdoch, then all three statements are false, which is also impossible, as one of the three defendants is a knight. Therefore, Murdoch must be $B$. | B | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,526 |
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