FORK IN THE ROAD - LOGIC QUESTIONS
You are travelling down a country lane to a distant village. You reach a fork in the road and find a pair of identical twin sisters standing there.
- One standing on the road to village and the other standing on the road to neverland (of course, you don't know or see where each road leads).
- One of the sisters always tells the truth and the other always lies (of course, you don't know who is lying).
- Both sisters know where the roads go.
If you are allowed to ask only one question to one of the sisters to find the correct road to the village, what is your question?
ANS=
This is one of the most famous logic problems which can be solved by using classic logic operations. You may have heard a few variations of this puzzle before (eg. 2 doors - 1 to heaven and 1 to hell) but still, it's one of the best brain teasers.
There are a few types of logic questions:
- Indirect question: "Hello there beauty, what would your sister say, if I asked her where this road leads?" The answer is always negated.
- Tricky question: "Excuse me lady, does a truth telling person stand on the road to the village?" The answer will be YES, if I am asking a truth teller who is standing at the road to village, or if I am asking a liar standing again on the same road. So I can go that way. A similar deduction can be made for negative answer.
- Complicated question: "Hey you, what would you say, if I asked you ...?" A truth teller is clear, but a liar should lie. However, she is forced by the question to lie two times and thus speak the truth.
- One standing on the road to village and the other standing on the road to neverland (of course, you don't know or see where each road leads).
- One of the sisters always tells the truth and the other always lies (of course, you don't know who is lying).
- Both sisters know where the roads go.
ANS=
This is one of the most famous logic problems which can be solved by using classic logic operations. You may have heard a few variations of this puzzle before (eg. 2 doors - 1 to heaven and 1 to hell) but still, it's one of the best brain teasers.
There are a few types of logic questions:
There are a few types of logic questions:
- Indirect question: "Hello there beauty, what would your sister say, if I asked her where this road leads?" The answer is always negated.
- Tricky question: "Excuse me lady, does a truth telling person stand on the road to the village?" The answer will be YES, if I am asking a truth teller who is standing at the road to village, or if I am asking a liar standing again on the same road. So I can go that way. A similar deduction can be made for negative answer.
- Complicated question: "Hey you, what would you say, if I asked you ...?" A truth teller is clear, but a liar should lie. However, she is forced by the question to lie two times and thus speak the truth.
HONESTANTS AND SWINDLECANTS I
There are two kinds of people on a mysterious island. There are so-called Honestants who speak always the truth, and the others are Swindlecants who always lie.
Three fellows (A, B and C) are having a quarrel at the market. A gringo goes by and asks the A fellow: "Are you an Honestant or a Swindlecant?" The answer is incomprehensible so the gringo gives another quite logical question to B: "What did A say?" B answers: "A said that he is a Swindlecant." And to that says the fellow C: "Do not believe B, he is lying!"
Who is B and C?
ANS=
It is impossible that any inhabitant of such an island says: "I am a liar." An honestant would thus be lying and a swindlecant would be speaking truth. So B must have been lying and therefore he is a swindlecant. And that means that C was right saying B is lying - so C is an honestant. However, it is not clear what is A.
Three fellows (A, B and C) are having a quarrel at the market. A gringo goes by and asks the A fellow: "Are you an Honestant or a Swindlecant?" The answer is incomprehensible so the gringo gives another quite logical question to B: "What did A say?" B answers: "A said that he is a Swindlecant." And to that says the fellow C: "Do not believe B, he is lying!"
Who is B and C?
ANS=
It is impossible that any inhabitant of such an island says: "I am a liar." An honestant would thus be lying and a swindlecant would be speaking truth. So B must have been lying and therefore he is a swindlecant. And that means that C was right saying B is lying - so C is an honestant. However, it is not clear what is A.
HONESTANTS AND SWINDLECANTS II
Afterwards he meets another two aborigines. One says: "I am a Swindlecant or the other one is an Honestant."
Who are they?
ANS=
Logical disjunction is a statement "P or Q". Such a disjunction is false if both P and Q are false. In all other cases it is true. Note that in everyday language, use of the word "or" can sometimes mean "either, but not both" (e.g., "would you like tea or coffee?"). In logic, this is called an "exclusive disjunction" or "exclusive or" (xor).
So if A was a swindlecant, then his statement would be false (thus A would have to be an honestant and B would have to be a swindlecant). However, that would cause a conflict which implicates that A must be an honestant. In that case at least one part of his statement is true and as it can't be the first one, B must be an honestant, too.
Who are they?
ANS=
Logical disjunction is a statement "P or Q". Such a disjunction is false if both P and Q are false. In all other cases it is true. Note that in everyday language, use of the word "or" can sometimes mean "either, but not both" (e.g., "would you like tea or coffee?"). In logic, this is called an "exclusive disjunction" or "exclusive or" (xor).
So if A was a swindlecant, then his statement would be false (thus A would have to be an honestant and B would have to be a swindlecant). However, that would cause a conflict which implicates that A must be an honestant. In that case at least one part of his statement is true and as it can't be the first one, B must be an honestant, too.
So if A was a swindlecant, then his statement would be false (thus A would have to be an honestant and B would have to be a swindlecant). However, that would cause a conflict which implicates that A must be an honestant. In that case at least one part of his statement is true and as it can't be the first one, B must be an honestant, too.
HONESTANTS AND SWINDLECANTS III
Our gringo displeased the sovereign with his intrusive questions and was condemned to death. But there was also a chance to save himself by solving the following logic problem. The gringo was shown two doors - one leading to a scaffold and the second one to freedom (both doors were the same) and only the door guards knew what was behind the doors. The sovereign let the gringo put one question to one guard. And because the sovereign was an honest man he warned that one guard is a Swindlecant.
What logic question can save the gringo's life?
You probably remember the answer from the very first problem on this page, don't you :-)
ANS=
There are a few types of questions:
- Indirect question: "Hey you, what would the other guard say, if I asked him where this door leads?" The answer is always negated.
- Tricky question: "Hey you, does an honestant stand at the door to freedom?" The answer will be YES, if I am asking an honestant who is standing at the door to freedom, or if I am asking a swindlecant standing again at the same door. So I can walk through the door. A similar deduction can be made for negative answer.
- Complicated question: "Hey you, what would you say, if I asked you ...?" An honestant is clear, but a swindlecant should lie. However, he is forced by the question to lie two times and thus speak the truth.
What logic question can save the gringo's life?
You probably remember the answer from the very first problem on this page, don't you :-)
ANS=
There are a few types of questions:
- Indirect question: "Hey you, what would the other guard say, if I asked him where this door leads?" The answer is always negated.
- Tricky question: "Hey you, does an honestant stand at the door to freedom?" The answer will be YES, if I am asking an honestant who is standing at the door to freedom, or if I am asking a swindlecant standing again at the same door. So I can walk through the door. A similar deduction can be made for negative answer.
- Complicated question: "Hey you, what would you say, if I asked you ...?" An honestant is clear, but a swindlecant should lie. However, he is forced by the question to lie two times and thus speak the truth.
HONESTANTS AND SWINDLECANTS IV
Our gringo was lucky and survived. On his way to the pub he met three aborigines. One made this statement: "We are all Swindlecants." The second one concluded: "Just one of us is an honest man."
Who are they?
ANS=
The first one must be a swindlecant (otherwise he would bring himself into a liar paradox), and so (knowing that the first one is lying) there must be at least one honestant among them. If the second one is lying, then (as the first one stated) the third one is an honestant, but that would make the second one speak the truth. So the second one is an honestant and C is a swindlecant.
Our gringo was lucky and survived. On his way to the pub he met three aborigines. One made this statement: "We are all Swindlecants." The second one concluded: "Just one of us is an honest man."Who are they?
ANS=
The first one must be a swindlecant (otherwise he would bring himself into a liar paradox), and so (knowing that the first one is lying) there must be at least one honestant among them. If the second one is lying, then (as the first one stated) the third one is an honestant, but that would make the second one speak the truth. So the second one is an honestant and C is a swindlecant.
HONESTANTS AND SWINDLECANTS V
In the pub the gringo met a funny guy who said: "If my wife is an Honestant, then I am Swindlecant."
Who is this couple?
ANS=
It is important to explore the statement as a whole. In this logical conditional ("if-then" statement) p is a hypothesis (or antecedent) and q is a conclusion (or consequent).
It is obvious, that the husband is not a Swindlecant, because in that case one part of the statement (Q) " ... then I am Swindlecant." would have to be a lie, which is a conflict. And since A is an Honestant, the whole statement is true.
If his wife was an Honestant too, then the second part of statement (Q) " ... then I am Swindlecant." would have to be true, which is a conflict again. Therefore the man is an Honestant and his wife is a Swindlecant. Or is it a paradox? Think about it.
Who is this couple?
ANS=
It is important to explore the statement as a whole. In this logical conditional ("if-then" statement) p is a hypothesis (or antecedent) and q is a conclusion (or consequent).
It is obvious, that the husband is not a Swindlecant, because in that case one part of the statement (Q) " ... then I am Swindlecant." would have to be a lie, which is a conflict. And since A is an Honestant, the whole statement is true.
If his wife was an Honestant too, then the second part of statement (Q) " ... then I am Swindlecant." would have to be true, which is a conflict again. Therefore the man is an Honestant and his wife is a Swindlecant. Or is it a paradox? Think about it.
It is obvious, that the husband is not a Swindlecant, because in that case one part of the statement (Q) " ... then I am Swindlecant." would have to be a lie, which is a conflict. And since A is an Honestant, the whole statement is true.
If his wife was an Honestant too, then the second part of statement (Q) " ... then I am Swindlecant." would have to be true, which is a conflict again. Therefore the man is an Honestant and his wife is a Swindlecant. Or is it a paradox? Think about it.
HONESTANTS AND SWINDLECANTS VI
When the gringo wanted to pay and leave the pub, the bartender told him how much his drink costed. It was quite expensive, so he asked the bartender if he spoke the truth. But the gringo did not hear the whispered answer so he questioned a man sitting next to him about it. And the man said: "The bartender said yes, but he is a big liar."
Who are they?
ANS=
This one seems not clear to me. However, the bartender and the man sitting next to the gringo must be one honestant and one swindlecant (not knowing who is who).
1. the bartender must have said: "Yes, I speak the truth" (no matter who he is)
2. the man sitting next to gringo said: "The bartender said yes, but he is a big liar.", which is true only if BOTH parts of the sentence are true
if it's true - the man is an honestant and the bartender a swindlecant,
if it's false = "he is a big liar" is false - bartender is an honestant and the man is a swindlecant.
Who are they?
ANS=
This one seems not clear to me. However, the bartender and the man sitting next to the gringo must be one honestant and one swindlecant (not knowing who is who).
1. the bartender must have said: "Yes, I speak the truth" (no matter who he is)
2. the man sitting next to gringo said: "The bartender said yes, but he is a big liar.", which is true only if BOTH parts of the sentence are true
if it's true - the man is an honestant and the bartender a swindlecant,
if it's false = "he is a big liar" is false - bartender is an honestant and the man is a swindlecant.
1. the bartender must have said: "Yes, I speak the truth" (no matter who he is)
2. the man sitting next to gringo said: "The bartender said yes, but he is a big liar.", which is true only if BOTH parts of the sentence are true
if it's true - the man is an honestant and the bartender a swindlecant,
if it's false = "he is a big liar" is false - bartender is an honestant and the man is a swindlecant.
HONESTANTS AND SWINDLECANTS VII
Going out of the pub, the gringo heard about a fantastic buried treasure. He wanted to be sure so he asked another man who replied:
"On this island is a treasure, only if I am an honest man."
So shall he go and find the treasure?
ANS=
It is important to explore the statement as a whole. If the man is an Honestant, then the whole statement must be true. One part of it, where he said that he is an honest man is true then and so the other part (about the treasure) must be true, too. However, if he is a Swindlecant, the whole statement is a lie. The part mentioning that he is an honest man is in that case of course a lie. Thus the other part must be truth. So there must be a treasure on the island, no matter what kind of man said the sentence.
"On this island is a treasure, only if I am an honest man."
So shall he go and find the treasure?
ANS=
It is important to explore the statement as a whole. If the man is an Honestant, then the whole statement must be true. One part of it, where he said that he is an honest man is true then and so the other part (about the treasure) must be true, too. However, if he is a Swindlecant, the whole statement is a lie. The part mentioning that he is an honest man is in that case of course a lie. Thus the other part must be truth. So there must be a treasure on the island, no matter what kind of man said the sentence.
HONESTANTS AND SWINDLECANTS VIII
Thinking about the treasure, the gringo forgot what day it was, so he asked four aborigines and got these answers:
A: Yesterday was Wednesday.
B: Tomorrow will be Sunday.
C: Today is Friday.
D: The day before yesterday was Thursday.
Because everything you need to know is how many people lied, I will not tell. What day of the week was it?
ANS=
The important thing was what we did not need to know. So if we knew how many people lied we would know the answer. And one more thing - B and D said the same.
If all of them lied, there would be 4 possible days to choose from (which one is not clear).
If only one of them spoke the truth, it could be A or C, so 2 possible days (not clear again).
If two of them were honest, it would have to be B and D saying that it was Saturday.
Neither 3 nor all 4 could have been honest because of an obvious conflict.
So it was Saturday.
A: Yesterday was Wednesday.
B: Tomorrow will be Sunday.
C: Today is Friday.
D: The day before yesterday was Thursday.
Because everything you need to know is how many people lied, I will not tell. What day of the week was it?
ANS=
The important thing was what we did not need to know. So if we knew how many people lied we would know the answer. And one more thing - B and D said the same.
If all of them lied, there would be 4 possible days to choose from (which one is not clear).
If only one of them spoke the truth, it could be A or C, so 2 possible days (not clear again).
If two of them were honest, it would have to be B and D saying that it was Saturday.
Neither 3 nor all 4 could have been honest because of an obvious conflict.
So it was Saturday.
If all of them lied, there would be 4 possible days to choose from (which one is not clear).
If only one of them spoke the truth, it could be A or C, so 2 possible days (not clear again).
If two of them were honest, it would have to be B and D saying that it was Saturday.
Neither 3 nor all 4 could have been honest because of an obvious conflict.
So it was Saturday.
HONESTANTS AND SWINDLECANTS IX
After a hard day the gringo wanted some time to relax. But a few minutes later two aborigines wanted to talk to him. To make things clear, the gringo asked: "Is at least one of you an honestant?" After the answer, there was no doubt.
Who are they and who answered?
ANS=
If the aborigine answered "Yes.", the gringo would not have been able to identify them. That means, the answer had to be "No.", and the one who said that was a liar and the other one was an honest man.
Who are they and who answered?
ANS=
If the aborigine answered "Yes.", the gringo would not have been able to identify them. That means, the answer had to be "No.", and the one who said that was a liar and the other one was an honest man.
HONESTANTS AND SWINDLECANTS X
There was a girl on this island, and everybody wanted her. However, she wanted just a rich swindlecant. If you were a rich swindlecant, how would you convince her saying only one sentence? And what if she wanted a rich honestant (and if you were one). Let us assume for this logic problem that there are only rich or poor people on the island.
ANS=
"I am a poor swindlecant." An honestant can not say such a sentence, so it is a lie. And that's why only a rich swindlecant can say that.
"I am not a poor honestant." A swindlecant can not say that, because it would be true. And that's why an honestant who is not poor (a rich one) said that.
ANS=
"I am a poor swindlecant." An honestant can not say such a sentence, so it is a lie. And that's why only a rich swindlecant can say that.
"I am not a poor honestant." A swindlecant can not say that, because it would be true. And that's why an honestant who is not poor (a rich one) said that.
"I am not a poor honestant." A swindlecant can not say that, because it would be true. And that's why an honestant who is not poor (a rich one) said that.
LOGIC PROBLEMS IN THE COURT OF LAW I
And now a few cases from the island of honestants and swindlecants. A prisoner at the bar was allowed to say one sentence to defend himself. After a while he said: "A swindlecant committed the crime."
Did it rescue him?
ANS=
Yes, the statement helped him. If he is an honestant, then a swindlecant committed the crime. If he is a swindlecant, then his statement points to an honestant who is guilty. Thus he is again innocent regarding the statement.
And now a few cases from the island of honestants and swindlecants. A prisoner at the bar was allowed to say one sentence to defend himself. After a while he said: "A swindlecant committed the crime."Did it rescue him?
ANS=
Yes, the statement helped him. If he is an honestant, then a swindlecant committed the crime. If he is a swindlecant, then his statement points to an honestant who is guilty. Thus he is again innocent regarding the statement.
LOGIC PROBLEMS IN THE COURT OF LAW II
A man accused of a crime, hired an attorney whose statements were always admitted by the court as undisputable truth. The following exchange took place in court.
Prosecutor: "If the accused committed the crime, he had an accomplice."
Defender: "That is not true!"
Did the attorney help his client?
ANS=
The statement of plaintiff is a lie only if the hypothesis (or antecedent) is true and conclusion (or consequent) is not true. So the solicitor did not help his client at all. He actually said that his client was guilty and there was no accomplice.
Prosecutor: "If the accused committed the crime, he had an accomplice."
Defender: "That is not true!"
Did the attorney help his client?
ANS=
The statement of plaintiff is a lie only if the hypothesis (or antecedent) is true and conclusion (or consequent) is not true. So the solicitor did not help his client at all. He actually said that his client was guilty and there was no accomplice.
LOGIC PROBLEMS IN THE COURT OF LAW III
You live on an island where there are only two kinds of people: the ones who always tell the truth (truth tellers) and those who always lie (liars). You are accused of crime and brought before the court, where you are allowed to speak only one sentence in your defense. What do you say in each of the following situations?
- If you were a liar (the court does not know that) and you were innocent. And it is an established fact that a liar committed the crime.
- Same situation as above, but you are the one who committed the crime.
- If you were a truth teller (the court does not know that) and you were innocent. And it is an established fact that a truth teller committed the crime.
- If you were innocent and it is an established fact that the crime was not committed by a "normal" person. Normal people are that new immigrant group who sometimes lie and sometimes speak the truth. What sentence, no matter whether you were a truth teller, liar, or normal, can prove your innocence?
ANS=
1. "I did it - I am guilty."
2. There is no such sentence.
3. "I am innocent."
4. "Either I am an honestant and innocent, or I am a swindlecant and guilty." = "I am either an innocent honestant, or a guilty swindlecant." The court could think this way:
4.1 If he is an honestant, then his statement is true and he is innocent.
4.2 If he is a swindlecant, then his statement is a lie and he is neither an innocent honestant nor a guilty swindlecant. This means that he is an innocent swindlecant.
4.3 If he is normal, then he is innocent since a normal man couldn't have done that.
- If you were a liar (the court does not know that) and you were innocent. And it is an established fact that a liar committed the crime.
- Same situation as above, but you are the one who committed the crime.
- If you were a truth teller (the court does not know that) and you were innocent. And it is an established fact that a truth teller committed the crime.
- If you were innocent and it is an established fact that the crime was not committed by a "normal" person. Normal people are that new immigrant group who sometimes lie and sometimes speak the truth. What sentence, no matter whether you were a truth teller, liar, or normal, can prove your innocence?
ANS=
1. "I did it - I am guilty."
2. There is no such sentence.
3. "I am innocent."
4. "Either I am an honestant and innocent, or I am a swindlecant and guilty." = "I am either an innocent honestant, or a guilty swindlecant." The court could think this way:
4.1 If he is an honestant, then his statement is true and he is innocent.
4.2 If he is a swindlecant, then his statement is a lie and he is neither an innocent honestant nor a guilty swindlecant. This means that he is an innocent swindlecant.
4.3 If he is normal, then he is innocent since a normal man couldn't have done that.
2. There is no such sentence.
3. "I am innocent."
4. "Either I am an honestant and innocent, or I am a swindlecant and guilty." = "I am either an innocent honestant, or a guilty swindlecant." The court could think this way:
4.1 If he is an honestant, then his statement is true and he is innocent.
4.2 If he is a swindlecant, then his statement is a lie and he is neither an innocent honestant nor a guilty swindlecant. This means that he is an innocent swindlecant.
4.3 If he is normal, then he is innocent since a normal man couldn't have done that.
PANDORA'S BOX I
Once upon a time, there was a girl named Pandora, who wanted a bright groom so she made up a few logic problems for the wannabe. This is one of them.
Based upon the inscriptions on the boxes (none or just one of them is true), choose one box where the wedding ring is hidden.
Golden box
The ring is in this box.
Silver box
The ring is not in this box.
Lead box
The ring is not in the golden box.
ANS=
The given conditions indicate that only the inscription on the lead box is true. So the ring is in the silver box.
Based upon the inscriptions on the boxes (none or just one of them is true), choose one box where the wedding ring is hidden.
Golden box
The ring is in this box.
The ring is in this box.
Silver box
The ring is not in this box.
The ring is not in this box.
Lead box
The ring is not in the golden box.
The ring is not in the golden box.
ANS=
The given conditions indicate that only the inscription on the lead box is true. So the ring is in the silver box.
PANDORA'S BOX II
And here is the second test. At least one inscription is true and at least one is false. Which means the ring is in the...
Golden box
The ring is not in the silver box.
Silver box
The ring is not in this box.
Lead box
The ring is in this box.
ANS=
The ring must be in the golden box, otherwise all the inscriptions would be either true or false.
Golden box
The ring is not in the silver box.
The ring is not in the silver box.
Silver box
The ring is not in this box.
The ring is not in this box.
Lead box
The ring is in this box.
The ring is in this box.
ANS=
The ring must be in the golden box, otherwise all the inscriptions would be either true or false.
LION AND UNICORN I
Alice came across a lion and a unicorn in a forest of forgetfulness. Those two are strange beings. The lion lies every Monday, Tuesday and Wednesday and the other days he speaks the truth. The unicorn lies on Thursdays, Fridays and Saturdays, however the other days of the week he speaks the truth.
Lion: Yesterday I was lying.
Unicorn: So was I.
Which day did they say that?
ANS=
As there is no day when both of the beings would be lying, at least one of them must have spoken the truth. They both speak the truth only on Sunday. However, the Lion would then be lying in his statement, so it couldn't be said on Sunday. So exactly one of them lied.
If the Unicorn was honest, then it would have to be Sunday - but previously we proved this wrong. Thus only the Lion spoke the truth when he met Alice on Thursday and spoke with the Unicorn about Wednesday.
Lion: Yesterday I was lying.
Unicorn: So was I.
Which day did they say that?
ANS=
As there is no day when both of the beings would be lying, at least one of them must have spoken the truth. They both speak the truth only on Sunday. However, the Lion would then be lying in his statement, so it couldn't be said on Sunday. So exactly one of them lied.
If the Unicorn was honest, then it would have to be Sunday - but previously we proved this wrong. Thus only the Lion spoke the truth when he met Alice on Thursday and spoke with the Unicorn about Wednesday.
If the Unicorn was honest, then it would have to be Sunday - but previously we proved this wrong. Thus only the Lion spoke the truth when he met Alice on Thursday and spoke with the Unicorn about Wednesday.
LION AND UNICORN II
Lion said: Yesterday I was lying and two days after tomorrow I will be lying again.
Which day did he say that?
ANS=
This conjunction is true only if both parts are true. The first part is true only on Thursday, but the second part is a lie then (Sunday is not a lying day of the Lion). So the whole statement can never be true (at least one part is not true). Therefore the Lion could have made the statement on Monday, on Tuesday and even on Wednesday.
Which day did he say that?
ANS=
This conjunction is true only if both parts are true. The first part is true only on Thursday, but the second part is a lie then (Sunday is not a lying day of the Lion). So the whole statement can never be true (at least one part is not true). Therefore the Lion could have made the statement on Monday, on Tuesday and even on Wednesday.
ISLAND BAAL
There are people and strange monkeys on this island, and you can not tell who is who. They speak either only the truth or only lies.
Who are the following two guys?
A: B is a lying monkey. I am human.
B: A is telling the truth.
ANS=
Conjunction used by A is true only if both parts are true. Under the assumption that B is an honest man, then A would be honest too (B says so) and so B would be a liar as A said, which would be a conflict. So B is a liar. And knowing that, B actually said that A is a liar, too. First statement of A is thus a lie and B is not a lying monkey. However, B is lying which means he is not a monkey. B is a lying man. The second statement of A indicates that A is a monkey - so A is a lying monkey.
Who are the following two guys?
A: B is a lying monkey. I am human.
B: A is telling the truth.
ANS=
Conjunction used by A is true only if both parts are true. Under the assumption that B is an honest man, then A would be honest too (B says so) and so B would be a liar as A said, which would be a conflict. So B is a liar. And knowing that, B actually said that A is a liar, too. First statement of A is thus a lie and B is not a lying monkey. However, B is lying which means he is not a monkey. B is a lying man. The second statement of A indicates that A is a monkey - so A is a lying monkey.
TRUTH, LIE AND WISDOM
Three goddesses were sitting in an old Indian temple. Their names were Truth (always telling the truth), Lie (always lying) and Wisdom (sometimes lying). A visitor asked the one on the left: "Who is sitting next to you?"
"Truth," she answered.
Then he asked the one in the middle: "Who are you?"
"Wisdom."
Lastly, he asked the one on the right: "Who is your neighbor?"
"Lie," she replied.
And then it became clear who is who.
ANS=
Let's assign a letter to each goddess. We get these sentences.
1. A says: B is Truth.
2. B says: I am Wisdom.
3. C says: B is Lie.
First sentence hints that A is not Truth. Second sentence is not said by Truth either, so C is Truth. Thus the third sentence is true. B is Lie and A is Wisdom.
"Truth," she answered.
Then he asked the one in the middle: "Who are you?"
"Wisdom."
Lastly, he asked the one on the right: "Who is your neighbor?"
"Lie," she replied.
And then it became clear who is who.
ANS=
Let's assign a letter to each goddess. We get these sentences.
1. A says: B is Truth.
2. B says: I am Wisdom.
3. C says: B is Lie.
First sentence hints that A is not Truth. Second sentence is not said by Truth either, so C is Truth. Thus the third sentence is true. B is Lie and A is Wisdom.
1. A says: B is Truth.
2. B says: I am Wisdom.
3. C says: B is Lie.
First sentence hints that A is not Truth. Second sentence is not said by Truth either, so C is Truth. Thus the third sentence is true. B is Lie and A is Wisdom.
IN THE ALPS
Three tourists have an argument regarding the way they should go. Hans says that Emanuel lies. Emanuel claims that Hans and Philip speak the same, only doesn't know whether truth or lie.
So who is lying for sure?
ANS
The only one who is lying for sure is Philip. Hans speaks probably the truth and Emanuel lies. It can be also the other way, but since Hans expressed himself before Emanuel did, then Emanuel's remark (that he does not know whether Hans is lying) is not true.
Three tourists have an argument regarding the way they should go. Hans says that Emanuel lies. Emanuel claims that Hans and Philip speak the same, only doesn't know whether truth or lie.
So who is lying for sure?
So who is lying for sure?
ANS
The only one who is lying for sure is Philip. Hans speaks probably the truth and Emanuel lies. It can be also the other way, but since Hans expressed himself before Emanuel did, then Emanuel's remark (that he does not know whether Hans is lying) is not true.
COINS
Imagine there are 3 coins on the table: gold, silver, and copper. If you make a truthful statement, you will get one coin. If you make a false statement, you will get nothing.
What sentence can guarantee you getting the gold coin?
ANS
"You will give me neither copper nor silver coin." If it is true, then I have to get the gold coin. If it is a lie, then the negation must be true, so "you give me either copper or silver coin", which would break the given conditions that you get no coin when lying. So the first sentence must be tr
Imagine there are 3 coins on the table: gold, silver, and copper. If you make a truthful statement, you will get one coin. If you make a false statement, you will get nothing.
What sentence can guarantee you getting the gold coin?
What sentence can guarantee you getting the gold coin?
ANS
"You will give me neither copper nor silver coin." If it is true, then I have to get the gold coin. If it is a lie, then the negation must be true, so "you give me either copper or silver coin", which would break the given conditions that you get no coin when lying. So the first sentence must be tr
SLIM LOVER
Something to relax. A slim young man asked a girl on a date:
"I say something. If it is truthful, will you give me your photo?"
"Yes," replied miss.
"And if it is a lie, do not give me your photograph. Would you promise that?"
The girl agreed. Then the chap said such a sentence, that after a little while of thinking she realized, that if she wanted to honor her promise, she wouldn't have to give him a photo but a kiss.
What would you say (if you were him) to be kissed and so on?
Ans
You could say for instance this sentence: "You will give me neither your photo nor a kiss."
Something to relax. A slim young man asked a girl on a date:
"I say something. If it is truthful, will you give me your photo?"
"Yes," replied miss.
"And if it is a lie, do not give me your photograph. Would you promise that?"
The girl agreed. Then the chap said such a sentence, that after a little while of thinking she realized, that if she wanted to honor her promise, she wouldn't have to give him a photo but a kiss.
What would you say (if you were him) to be kissed and so on?
"I say something. If it is truthful, will you give me your photo?"
"Yes," replied miss.
"And if it is a lie, do not give me your photograph. Would you promise that?"
The girl agreed. Then the chap said such a sentence, that after a little while of thinking she realized, that if she wanted to honor her promise, she wouldn't have to give him a photo but a kiss.
What would you say (if you were him) to be kissed and so on?
Ans
You could say for instance this sentence: "You will give me neither your photo nor a kiss."
BEAR
Albert Einstein allegedly made this riddle for his scholars.
A fellow encountered a bear in a wasteland. There was nobody else there. Both were frightened and ran away. Fellow to the north, bear to the west. Suddenly the fellow stopped, aimed his gun to the south and shot the bear. What color was the bear?
If you don't know, this may help you: if the bear ran about 3.14 times faster than the fellow (still westwards), the fellow could have shot straight in front of him, however for the booty he would have to go to the south.
ANS=
It all happened on the North Pole. When the man shot, he must have been right on the North Pole. Getting it? So it makes sense to assume that the only color the bear could be was WHITE.
So this is it. I've heard another logical solutions (even that there are no bears neither on the North nor on the South Pole), but this one presented makes sense to me. And what about you?
Albert Einstein allegedly made this riddle for his scholars.A fellow encountered a bear in a wasteland. There was nobody else there. Both were frightened and ran away. Fellow to the north, bear to the west. Suddenly the fellow stopped, aimed his gun to the south and shot the bear. What color was the bear?
If you don't know, this may help you: if the bear ran about 3.14 times faster than the fellow (still westwards), the fellow could have shot straight in front of him, however for the booty he would have to go to the south.
ANS=
It all happened on the North Pole. When the man shot, he must have been right on the North Pole. Getting it? So it makes sense to assume that the only color the bear could be was WHITE.
So this is it. I've heard another logical solutions (even that there are no bears neither on the North nor on the South Pole), but this one presented makes sense to me. And what about you?
So this is it. I've heard another logical solutions (even that there are no bears neither on the North nor on the South Pole), but this one presented makes sense to me. And what about you?
NEIGHBORS
This is another example of Einstein's riddles. It is said that this quiz was made up by the famous physicist and according to him 98% will not solve it.
There is a row of five different color houses. Each house is occupied by a man of different nationality. Each man has a different pet, prefers a different drink, and smokes different brand of cigarettes.
- The Brit lives in the Red house.
- The Swede keeps dogs as pets.
- The Dane drinks tea.
- The Green house is next to the White house, on the left.
- The owner of the Green house drinks coffee.
- The person who smokes Pall Mall rears birds.
- The owner of the Yellow house smokes Dunhill.
- The man living in the centre house drinks milk.
- The Norwegian lives in the first house.
- The man who smokes Blends lives next to the one who keeps cats.
- The man who keeps horses lives next to the man who smokes Dunhill.
- The man who smokes Blue Master drinks beer.
- The German smokes Prince.
- The Norwegian lives next to the Blue house.
- The man who smokes Blends has a neighbour who drinks water.
Who has fish at home? Are you one of the 2%?
ANS=
There is a row of five different color houses. Each house is occupied by a man of different nationality. Each man has a different pet, prefers a different drink, and smokes different brand of cigarettes.
- The Brit lives in the Red house.
- The Swede keeps dogs as pets.
- The Dane drinks tea.
- The Green house is next to the White house, on the left.
- The owner of the Green house drinks coffee.
- The person who smokes Pall Mall rears birds.
- The owner of the Yellow house smokes Dunhill.
- The man living in the centre house drinks milk.
- The Norwegian lives in the first house.
- The man who smokes Blends lives next to the one who keeps cats.
- The man who keeps horses lives next to the man who smokes Dunhill.
- The man who smokes Blue Master drinks beer.
- The German smokes Prince.
- The Norwegian lives next to the Blue house.
- The man who smokes Blends has a neighbour who drinks water.
ANS=
MEETING (MEET THIS CHALLENGE)
Another example of hard grid puzzles (just like Einstein's) was published in the QUIZ 11/1986.
Eight married couples meet to lend one another some books. Couples have the same surname, employment and car. Each couple has a favorite color. Furthermore we know the following facts:
- Daniella Black and her husband work as Shop-Assistants.
- The book "The Seadog" was brought by a couple who drive a Fiat and love the color red.
- Owen and his wife Victoria like the color brown.
- Stan Horricks and his wife Hannah like the color white.
- Jenny Smith and her husband work as Warehouse Managers and they drive a Wartburg.
- Monica and her husband Alexander borrowed the book "Grandfather Joseph".
- Mathew and his wife like the color pink and brought the book "Mulatka Gabriela".
- Irene and her husband Oto work as Accountants.
- The book "We Were Five" was borrowed by a couple driving a Trabant.
- The Cermaks are both Ticket-Collectors who brought the book "Shed Stoat".
- Mr and Mrs Kuril are both Doctors who borrowed the book "Slovacko Judge".
- Paul and his wife like the color green.
- Veronica Dvorak and her husband like the color blue.
- Rick and his wife brought the book "Slovacko Judge" and they drive a Ziguli.
- One couple brought the book "Dame Commissar" and borrowed the book "Mulatka Gabriela".
- The couple who drive a Dacia, love the color violet.
- The couple who work as Teachers borrowed the book "Dame Commissar".
- The couple who work as Agriculturalists drive a Moskvic.
- Pamela and her husband drive a Renault and brought the book "Grandfather Joseph".
- Pamela and her husband borrowed the book that Mr and Mrs Zajac brought.
- Robert and his wife like the color yellow and borrowed the book "The Modern Comedy".
- Mr and Mrs Swain work as Shoppers.
- "The Modern Comedy" was brought by a couple driving a Skoda.
Is it a problem to find out everything about everyone from this information?
ANS=
Eight married couples meet to lend one another some books. Couples have the same surname, employment and car. Each couple has a favorite color. Furthermore we know the following facts:
- Daniella Black and her husband work as Shop-Assistants.
- The book "The Seadog" was brought by a couple who drive a Fiat and love the color red.
- Owen and his wife Victoria like the color brown.
- Stan Horricks and his wife Hannah like the color white.
- Jenny Smith and her husband work as Warehouse Managers and they drive a Wartburg.
- Monica and her husband Alexander borrowed the book "Grandfather Joseph".
- Mathew and his wife like the color pink and brought the book "Mulatka Gabriela".
- Irene and her husband Oto work as Accountants.
- The book "We Were Five" was borrowed by a couple driving a Trabant.
- The Cermaks are both Ticket-Collectors who brought the book "Shed Stoat".
- Mr and Mrs Kuril are both Doctors who borrowed the book "Slovacko Judge".
- Paul and his wife like the color green.
- Veronica Dvorak and her husband like the color blue.
- Rick and his wife brought the book "Slovacko Judge" and they drive a Ziguli.
- One couple brought the book "Dame Commissar" and borrowed the book "Mulatka Gabriela".
- The couple who drive a Dacia, love the color violet.
- The couple who work as Teachers borrowed the book "Dame Commissar".
- The couple who work as Agriculturalists drive a Moskvic.
- Pamela and her husband drive a Renault and brought the book "Grandfather Joseph".
- Pamela and her husband borrowed the book that Mr and Mrs Zajac brought.
- Robert and his wife like the color yellow and borrowed the book "The Modern Comedy".
- Mr and Mrs Swain work as Shoppers.
- "The Modern Comedy" was brought by a couple driving a Skoda.
ANS=
SHIPS
The following grid puzzle might be easy.
There are 5 ships in a port.
- The Greek ship leaves at six and carries coffee.
- The ship in the middle has a black chimney.
- The English ship leaves at nine.
- The French ship with a blue chimney is to the left of a ship that carries coffee.
- To the right of the ship carrying cocoa is a ship going to Marseille.
- The Brazilian ship is heading for Manila.
- Next to the ship carrying rice is a ship with a green chimney.
- A ship going to Genoa leaves at five.
- The Spanish ship leaves at seven and is to the right of the ship going to Marseille.
- The ship with a red chimney goes to Hamburg.
- Next to the ship leaving at seven is a ship with a white chimney.
- The ship on the border carries corn.
- The ship with a black chimney leaves at eight.
- The ship carrying corn is anchored next to the ship carrying rice.
- The ship to Hamburg leaves at six.
Which ship goes to Port Said? Which ship carries tea?
ANS=
There are 5 ships in a port.
- The Greek ship leaves at six and carries coffee.
- The ship in the middle has a black chimney.
- The English ship leaves at nine.
- The French ship with a blue chimney is to the left of a ship that carries coffee.
- To the right of the ship carrying cocoa is a ship going to Marseille.
- The Brazilian ship is heading for Manila.
- Next to the ship carrying rice is a ship with a green chimney.
- A ship going to Genoa leaves at five.
- The Spanish ship leaves at seven and is to the right of the ship going to Marseille.
- The ship with a red chimney goes to Hamburg.
- Next to the ship leaving at seven is a ship with a white chimney.
- The ship on the border carries corn.
- The ship with a black chimney leaves at eight.
- The ship carrying corn is anchored next to the ship carrying rice.
- The ship to Hamburg leaves at six.
ANS=
GARDENS
Five friends have their gardens next to one another, where they grow three kinds of crops: fruits (apple, pear, nut, cherry), vegetables (carrot, parsley, gourd, onion) and flowers (aster, rose, tulip, lily).
- They grow 12 different varieties.
- Everybody grows exactly 4 different varieties
- Each variety is at least in one garden.
- Only one variety is in 4 gardens.
- Only in one garden are all 3 kinds of crops.
- Only in one garden are all 4 varieties of one kind of crops.
- Pears are only in the two border gardens.
- Paul's garden is in the middle with no lily.
- Aster grower doesn't grow vegetables.
- Rose grower doesn't grow parsley.
- Nuts grower has also gourd and parsley.
- In the first garden are apples and cherries.
- Only in two gardens are cherries.
- Sam has onions and cherries.
- Luke grows exactly two kinds of fruit.
- Tulips are only in two gardens.
- Apples are in a single garden.
- Only in one garden next to the Zick's is parsley.
- Sam's garden is not on the border.
- Hank grows neither vegetables nor asters.
- Paul has exactly three kinds of vegetable.
Who has which garden and what is grown where?
ANS=
- They grow 12 different varieties.
- Everybody grows exactly 4 different varieties
- Each variety is at least in one garden.
- Only one variety is in 4 gardens.
- Only in one garden are all 3 kinds of crops.
- Only in one garden are all 4 varieties of one kind of crops.
- Pears are only in the two border gardens.
- Paul's garden is in the middle with no lily.
- Aster grower doesn't grow vegetables.
- Rose grower doesn't grow parsley.
- Nuts grower has also gourd and parsley.
- In the first garden are apples and cherries.
- Only in two gardens are cherries.
- Sam has onions and cherries.
- Luke grows exactly two kinds of fruit.
- Tulips are only in two gardens.
- Apples are in a single garden.
- Only in one garden next to the Zick's is parsley.
- Sam's garden is not on the border.
- Hank grows neither vegetables nor asters.
- Paul has exactly three kinds of vegetable.
ANS=
1. WEIGHING IN A HARDER WAY
You've got 27 coins, each of them is 10g, except for 1. The 1 different coin is 9g or 11g (heavier, or lighter by 1g). You should use balance scale that compares what's in the two pans. You can get the answer by just comparing groups of coins.
What is the minimum number weighings that can always guarantee to determine the different coin.
ANS=
Separate the coins into 3 stacks of 9 (A, B, C). Weigh stack A against B and then A against C. Take the stack with the different weight (note lighter or heavier) and break it into 3 stacks of 3 (D, E, F). Weigh stack D against E. If D and E are equal, then F is the odd stack. If D and E are not equal, the lighter or heavier (based on the A, B, C comparison) is the odd stack. You now have three coins (G, H, I). Weigh G and H. If G equals H, then I is the odd and is lighter or heavier (based on the A, B, C comparison). If G and H are not equal, then the lighter or heavier (based on the A, B, C comparison) is the odd coin.
What is the minimum number weighings that can always guarantee to determine the different coin.
ANS=
Separate the coins into 3 stacks of 9 (A, B, C). Weigh stack A against B and then A against C. Take the stack with the different weight (note lighter or heavier) and break it into 3 stacks of 3 (D, E, F). Weigh stack D against E. If D and E are equal, then F is the odd stack. If D and E are not equal, the lighter or heavier (based on the A, B, C comparison) is the odd stack. You now have three coins (G, H, I). Weigh G and H. If G equals H, then I is the odd and is lighter or heavier (based on the A, B, C comparison). If G and H are not equal, then the lighter or heavier (based on the A, B, C comparison) is the odd coin.
2. HATS ON A DEATH ROW
You are one of 20 prisoners on death row with the execution date set for tomorrow. Your king is a ruthless man who likes to toy with his people's miseries. He comes to your cell today and tells you:
"I'm gonna give you prisoners a chance to go free tomorrow. You will all stand in a row (queue) before the executioner and we will put a hat on your head, either a red or a black one. Of course you will not be able to see the color of your own hat; you will only be able to see the prisoners in front of you with their hats on; you will not be allowed to look back or communicate together in any way (talking, touching.....).
The prisoner in the back will be able to see the 19 prisoners in front of him. The one in front of him will be able to see 18...
Starting with the last person in the row, the one who can see everybody in front of him, he will be asked a simple question: WHAT IS THE COLOR OF YOUR HAT?
He will be only allowed to answer "BLACK" or "RED". If he says anything else you will ALL be executed immediately.
If he guesses the right color of the hat on his head he is set free, otherwise he is put to death. And we move on to the one in front of him and ask him the same question and so on...
Well, good luck tomorrow, HA HA HA HA HA HA!"
Now since you all can communicate freely during the night, can you find a way to guarantee the freedom of some prisoners tomorrow? How many?
ANS=
First guy is a coin toss - let's wish him good luck.
His job is to establish the parity of black hats visible to him.
He says "Black" if he sees an odd number of black hats; "Red" otherwise.
By paying attention to what has been said, each prisoner will know his hat's color.
Example:
Second to speak hears "Black" and sees an even number of black hats.
He knows his hat is black [odd changed to even - must be his is black] and says "black".
Third guy has heard "black" and "black" and sees an even number of black hats.
He knows his hat is red [even stayed even - his hat can't be black] and says "red".
And so on, to the front of the line.
General algorithm:
The first time you hear "black", say to yourself "odd".
Each time your hear "black" after that, change the parity: "even", "odd", ... etc.
When it's your turn, if the black hats you see match the running parity, you're Red; Black otherwise.
Call out your color.
If you like this type of brain teasers, then surely check out other logic puzzles. Especially "Masters of Logic Puzzles" series at the bottom of that page might be of interest for you.
"I'm gonna give you prisoners a chance to go free tomorrow. You will all stand in a row (queue) before the executioner and we will put a hat on your head, either a red or a black one. Of course you will not be able to see the color of your own hat; you will only be able to see the prisoners in front of you with their hats on; you will not be allowed to look back or communicate together in any way (talking, touching.....).
The prisoner in the back will be able to see the 19 prisoners in front of him. The one in front of him will be able to see 18...
Starting with the last person in the row, the one who can see everybody in front of him, he will be asked a simple question: WHAT IS THE COLOR OF YOUR HAT?
He will be only allowed to answer "BLACK" or "RED". If he says anything else you will ALL be executed immediately.
If he guesses the right color of the hat on his head he is set free, otherwise he is put to death. And we move on to the one in front of him and ask him the same question and so on...
Well, good luck tomorrow, HA HA HA HA HA HA!"
Now since you all can communicate freely during the night, can you find a way to guarantee the freedom of some prisoners tomorrow? How many?
ANS=
First guy is a coin toss - let's wish him good luck.
His job is to establish the parity of black hats visible to him.
He says "Black" if he sees an odd number of black hats; "Red" otherwise.
By paying attention to what has been said, each prisoner will know his hat's color.
Example:
Second to speak hears "Black" and sees an even number of black hats.
He knows his hat is black [odd changed to even - must be his is black] and says "black".
Third guy has heard "black" and "black" and sees an even number of black hats.
He knows his hat is red [even stayed even - his hat can't be black] and says "red".
And so on, to the front of the line.
General algorithm:
The first time you hear "black", say to yourself "odd".
Each time your hear "black" after that, change the parity: "even", "odd", ... etc.
When it's your turn, if the black hats you see match the running parity, you're Red; Black otherwise.
Call out your color.
If you like this type of brain teasers, then surely check out other logic puzzles. Especially "Masters of Logic Puzzles" series at the bottom of that page might be of interest for you.
His job is to establish the parity of black hats visible to him.
He says "Black" if he sees an odd number of black hats; "Red" otherwise.
By paying attention to what has been said, each prisoner will know his hat's color.
Example:
Second to speak hears "Black" and sees an even number of black hats.
He knows his hat is black [odd changed to even - must be his is black] and says "black".
Third guy has heard "black" and "black" and sees an even number of black hats.
He knows his hat is red [even stayed even - his hat can't be black] and says "red".
And so on, to the front of the line.
General algorithm:
The first time you hear "black", say to yourself "odd".
Each time your hear "black" after that, change the parity: "even", "odd", ... etc.
When it's your turn, if the black hats you see match the running parity, you're Red; Black otherwise.
Call out your color.
If you like this type of brain teasers, then surely check out other logic puzzles. Especially "Masters of Logic Puzzles" series at the bottom of that page might be of interest for you.
3. THE LIAR, THE TRUTH TELLER AND THE RANDOM ANSWERER
There is a truth teller (always tells the truth), a liar (always lies), and one that sometimes answers truthfully and sometimes lies. Each man knows who is who. You may ask three yes or no question to determine who is who. Each time you ask a question, it must only be directed to one of the men (of your choice). You may ask the same question more than once, but of course it will count towards your total.
What are your questions and to whom will you ask them?
ANS=
the order of the men: TRL, TLR, LTR, LRT, RTL, RLT
There are 8 possible combinations of anwers for questions: TTT, TTL, TLT, TLL, LTT, LTL, LLT, LLL.
Theoretically it's possbile if you could figure out a way to get any of the 8 combinations of answers assigned to the states, but with the unreliability of Random's answers, I thought it was impossible. There is always a possiblity in any solution where Random will exactly mirror T or L for answers. He could always lie or always tell the truth and you can never tell when he is lying or telling the truth. This being given, I thought you can NEVER separate 6 distinct answers to apply to the 6 states, and therefore can never be sure who is who.
After a minute though, I saw through my own error in logic. I was always dealing with questions where T and L would give the same answer regardless of the order of the men. I saw that if you can get T and L to give a Yes/NO answer, then you can figure out where R's worthless answers are. The only way I saw to do this is to ask about the order of the men themselves.
So:
Ask #1 if L is standing on R's right arm (our left if they are facing us).
The answer gives you a split in the order they are standing:
If YES, then it has to be T telling the truth, L telling a lie, or one of R's worthless answers, so: TLR, LTR, or RTL, RLT.
If NO, then it has to be T telling the truth, L telling a lie, or R and his worthless answers, so: TRL, LRT, or RTL, RLT.
Now we know, based on the answer to #1 where to avoid R's worthless answers. We now ask T or L "Is T in the lineup?" If answer 1 was Yes, we ask person 2, if it was no we ask person 3.
The answer now will give us some more info. If it's Yes, it's T answering the truth, if it's no, it's L answering a lie. So based on who we asked, we now know:
Yes, Yes: Has to be LTR, or RTL
Yes, No: TLR, RLT
No, Yes: LRT, RLT
No, No: TRL, RTL
Now any question separating the two possiblities works - just make sure you are avoiding R's worthless answers.
For example:
Yes, Yes - ask #2 if #1 is L. (We know #2 is T and will tell the truth) - Yes = LTR, No = RTL
Yes, No, - ask #2 if #1 is T. ( We know #2 is L and will tell a lie) - Yes = RLT, No = TLR
No, Yes - ask # 3 if #1 is L. (We know #3 is T and will tell the truth) - Yes = LRT, No = RLT
No, No, - ask #3 if #1 is T. (We know #3 is L and will tell a lie) - Yes = RTL, No = TRL
So we have the order and know who is who.
There is a truth teller (always tells the truth), a liar (always lies), and one that sometimes answers truthfully and sometimes lies. Each man knows who is who. You may ask three yes or no question to determine who is who. Each time you ask a question, it must only be directed to one of the men (of your choice). You may ask the same question more than once, but of course it will count towards your total.What are your questions and to whom will you ask them?
ANS=
the order of the men: TRL, TLR, LTR, LRT, RTL, RLT
There are 8 possible combinations of anwers for questions: TTT, TTL, TLT, TLL, LTT, LTL, LLT, LLL.
Theoretically it's possbile if you could figure out a way to get any of the 8 combinations of answers assigned to the states, but with the unreliability of Random's answers, I thought it was impossible. There is always a possiblity in any solution where Random will exactly mirror T or L for answers. He could always lie or always tell the truth and you can never tell when he is lying or telling the truth. This being given, I thought you can NEVER separate 6 distinct answers to apply to the 6 states, and therefore can never be sure who is who.
After a minute though, I saw through my own error in logic. I was always dealing with questions where T and L would give the same answer regardless of the order of the men. I saw that if you can get T and L to give a Yes/NO answer, then you can figure out where R's worthless answers are. The only way I saw to do this is to ask about the order of the men themselves.
So:
Ask #1 if L is standing on R's right arm (our left if they are facing us).
The answer gives you a split in the order they are standing:
If YES, then it has to be T telling the truth, L telling a lie, or one of R's worthless answers, so: TLR, LTR, or RTL, RLT.
If NO, then it has to be T telling the truth, L telling a lie, or R and his worthless answers, so: TRL, LRT, or RTL, RLT.
Now we know, based on the answer to #1 where to avoid R's worthless answers. We now ask T or L "Is T in the lineup?" If answer 1 was Yes, we ask person 2, if it was no we ask person 3.
The answer now will give us some more info. If it's Yes, it's T answering the truth, if it's no, it's L answering a lie. So based on who we asked, we now know:
Yes, Yes: Has to be LTR, or RTL
Yes, No: TLR, RLT
No, Yes: LRT, RLT
No, No: TRL, RTL
Now any question separating the two possiblities works - just make sure you are avoiding R's worthless answers.
For example:
Yes, Yes - ask #2 if #1 is L. (We know #2 is T and will tell the truth) - Yes = LTR, No = RTL
Yes, No, - ask #2 if #1 is T. ( We know #2 is L and will tell a lie) - Yes = RLT, No = TLR
No, Yes - ask # 3 if #1 is L. (We know #3 is T and will tell the truth) - Yes = LRT, No = RLT
No, No, - ask #3 if #1 is T. (We know #3 is L and will tell a lie) - Yes = RTL, No = TRL
So we have the order and know who is who.
4. TRUTH IN PACKAGING
Three boxes are all labeled incorrectly, and you must get the labels right.
The labels on the boxes read as follows:
Box 1
nails
Box 2
screws
Box 3
nails and screws
To gain the information you need to move the labels to the correct boxes, you may remove a single item from one of the boxes. You may not look into the boxes, nor pick them up and shake them, etc.
Can this be done? If so, how? If not, why not?
ANS=
Remove an item from box 3.
The item tells you what label to put on box 3.
Move the nails and screws label to the box labeled with the other item, and its label to the remaining box.
Example: you remove a nail from box 3.
Move the label nails from box 1 to box 3.
You can't move the nails and screws label to box 1: that would be a swap, and all three labels must be corrected.
Move it instead to box 2, and the screws label to box 1.
[box 1] screws
[box 2] nails and screws
[box 3] nails
The labels on the boxes read as follows:
Box 1
nails
nails
Box 2
screws
screws
Box 3
nails and screws
To gain the information you need to move the labels to the correct boxes, you may remove a single item from one of the boxes. You may not look into the boxes, nor pick them up and shake them, etc.nails and screws
Can this be done? If so, how? If not, why not?
ANS=
Remove an item from box 3.
The item tells you what label to put on box 3.
Move the nails and screws label to the box labeled with the other item, and its label to the remaining box.
Example: you remove a nail from box 3.
Move the label nails from box 1 to box 3.
You can't move the nails and screws label to box 1: that would be a swap, and all three labels must be corrected.
Move it instead to box 2, and the screws label to box 1.
[box 1] screws
[box 2] nails and screws
[box 3] nails
The item tells you what label to put on box 3.
Move the nails and screws label to the box labeled with the other item, and its label to the remaining box.
Example: you remove a nail from box 3.
Move the label nails from box 1 to box 3.
You can't move the nails and screws label to box 1: that would be a swap, and all three labels must be corrected.
Move it instead to box 2, and the screws label to box 1.
[box 1] screws
[box 2] nails and screws
[box 3] nails
5. ONE GIRL - ONE BOY
Teanchi and Beanchi are a married couple (dont ask me who he is and who she is)! They have two kids, one of them is a girl. Assume safely that the probability of each gender is 1/2.
What is the probability that the other kid is also a girl?
Hint: It is not 1/2 as you would first think.
ANS=
Of course, it's not 1/2 else would make it a lousy puzzle.
Ans: 1/3
This is a famous question in understanding conditional probability, which simply means that given some information you might be able to get a better estimate.
The following are possible combinations of two children that form a sample space in any earthly family:
Girl - Girl
Girl - Boy
Boy - Girl
Boy - Boy
Since we know one of the children is a girl, we will drop the Boy-Boy possibility from the sample space.
This leaves only three possibilities, one of which is two girls. Hence the probability is 1/3
What is the probability that the other kid is also a girl?
Hint: It is not 1/2 as you would first think.
ANS=
Of course, it's not 1/2 else would make it a lousy puzzle.
Ans: 1/3
This is a famous question in understanding conditional probability, which simply means that given some information you might be able to get a better estimate.
The following are possible combinations of two children that form a sample space in any earthly family:
Girl - Girl
Girl - Boy
Boy - Girl
Boy - Boy
Since we know one of the children is a girl, we will drop the Boy-Boy possibility from the sample space.
This leaves only three possibilities, one of which is two girls. Hence the probability is 1/3
Ans: 1/3
This is a famous question in understanding conditional probability, which simply means that given some information you might be able to get a better estimate.
The following are possible combinations of two children that form a sample space in any earthly family:
Girl - Girl
Girl - Boy
Boy - Girl
Boy - Boy
Since we know one of the children is a girl, we will drop the Boy-Boy possibility from the sample space.
This leaves only three possibilities, one of which is two girls. Hence the probability is 1/3
6. WHATCHYA GONNA DO (2 GOATS AND A CAR)
You are on a game show and there are three doors. The presenter tells you that behind one of doors there is a car and behind the other two are goats. If you pick the car you win it. After you have picked a door the presenter opens a different door with a goat behind it, he then gives you the chance to change what door you open. What should you do?
Hint: It is not 1/2 as you would first think.
ANS=
Jennifer should switch. Contrary to what may seem intuitive, switching actually doubles her chances of winning the car.
This problem is just a re-wording of what is known as the Monty Hall Problem. The key to understanding it is that the host knows the locations of the car and goats. His knowledge changes his actions and thus affects the odds.
Here is a breakdown of all the possible scenarios that Jennifer faces and why Jennifer should switch:
Door #2 has goat B (probability 1:3) - MB shows goat A behind Door #1 (1:1) - the car is behind Door #3 (1:1) - switching wins the car - total chances (1:3 x 1:1 x 1:1 = 1:3)
Door #2 has the car (probability 1:3) - MB shows goat A behind Door #1 (1:2) - goat B is behind Door #3 (1:1) - not switching wins the car - total chances (1:3 x 1:2 x 1:1 = 1:6)
Door #2 has the car (probability 1:3) - MB shows goat B behind Door #1 (1:2) - goat A is behind Door #3 (1:1) - not switching wins the car - total chances (1:3 x 1:2 x 1:1 = 1:6)
There are (1:3 + 1:3 = 2:3) chances that switching will get Jennifer the car, and only (1:6 + 1:6 = 1:3) chances she would get the car by not switching. She should switch.
A more general presentation of the reasoning is this:
At the start of the game, there is a 2:3 chance that Jennifer will pick a door with a goat behind it. If she does, the host will reveal the other other goat, and switching doors will get Jennifer the car.
There is a 1:3 chance she will pick the car. The host will then reveal a goat. Switching would win Jennifer a good supply of Ch?vre (and the disdain of her neighbours).
So, 2 out of 3 times switching gets the car. Simple - unintuitive, but simple.
Why does the host's knowledge change the odds. Because he does not randomly select a door to open - he always opens a door with a goat. By doing this he reduced the possible scenarios for Jennifer to the four listed above. If he randomly picked, then Jenny's chances, if the show progressed as presented, would be 50/50. However, there would also be a 1:3 chance that MB would open the wrong door and reveal the car's location (followed by a 1:1 chance that MB would be sacked and re-runs of McGyver would fill out the remainder of the season!)
Hint: It is not 1/2 as you would first think.
ANS=
Jennifer should switch. Contrary to what may seem intuitive, switching actually doubles her chances of winning the car.
This problem is just a re-wording of what is known as the Monty Hall Problem. The key to understanding it is that the host knows the locations of the car and goats. His knowledge changes his actions and thus affects the odds.
Here is a breakdown of all the possible scenarios that Jennifer faces and why Jennifer should switch:
Door #2 has goat B (probability 1:3) - MB shows goat A behind Door #1 (1:1) - the car is behind Door #3 (1:1) - switching wins the car - total chances (1:3 x 1:1 x 1:1 = 1:3)
Door #2 has the car (probability 1:3) - MB shows goat A behind Door #1 (1:2) - goat B is behind Door #3 (1:1) - not switching wins the car - total chances (1:3 x 1:2 x 1:1 = 1:6)
Door #2 has the car (probability 1:3) - MB shows goat B behind Door #1 (1:2) - goat A is behind Door #3 (1:1) - not switching wins the car - total chances (1:3 x 1:2 x 1:1 = 1:6)
There are (1:3 + 1:3 = 2:3) chances that switching will get Jennifer the car, and only (1:6 + 1:6 = 1:3) chances she would get the car by not switching. She should switch.
A more general presentation of the reasoning is this:
At the start of the game, there is a 2:3 chance that Jennifer will pick a door with a goat behind it. If she does, the host will reveal the other other goat, and switching doors will get Jennifer the car.
There is a 1:3 chance she will pick the car. The host will then reveal a goat. Switching would win Jennifer a good supply of Ch?vre (and the disdain of her neighbours).
So, 2 out of 3 times switching gets the car. Simple - unintuitive, but simple.
Why does the host's knowledge change the odds. Because he does not randomly select a door to open - he always opens a door with a goat. By doing this he reduced the possible scenarios for Jennifer to the four listed above. If he randomly picked, then Jenny's chances, if the show progressed as presented, would be 50/50. However, there would also be a 1:3 chance that MB would open the wrong door and reveal the car's location (followed by a 1:1 chance that MB would be sacked and re-runs of McGyver would fill out the remainder of the season!)
This problem is just a re-wording of what is known as the Monty Hall Problem. The key to understanding it is that the host knows the locations of the car and goats. His knowledge changes his actions and thus affects the odds.
Here is a breakdown of all the possible scenarios that Jennifer faces and why Jennifer should switch:
Door #2 has goat B (probability 1:3) - MB shows goat A behind Door #1 (1:1) - the car is behind Door #3 (1:1) - switching wins the car - total chances (1:3 x 1:1 x 1:1 = 1:3)
Door #2 has the car (probability 1:3) - MB shows goat A behind Door #1 (1:2) - goat B is behind Door #3 (1:1) - not switching wins the car - total chances (1:3 x 1:2 x 1:1 = 1:6)
Door #2 has the car (probability 1:3) - MB shows goat B behind Door #1 (1:2) - goat A is behind Door #3 (1:1) - not switching wins the car - total chances (1:3 x 1:2 x 1:1 = 1:6)
There are (1:3 + 1:3 = 2:3) chances that switching will get Jennifer the car, and only (1:6 + 1:6 = 1:3) chances she would get the car by not switching. She should switch.
A more general presentation of the reasoning is this:
At the start of the game, there is a 2:3 chance that Jennifer will pick a door with a goat behind it. If she does, the host will reveal the other other goat, and switching doors will get Jennifer the car.
There is a 1:3 chance she will pick the car. The host will then reveal a goat. Switching would win Jennifer a good supply of Ch?vre (and the disdain of her neighbours).
So, 2 out of 3 times switching gets the car. Simple - unintuitive, but simple.
Why does the host's knowledge change the odds. Because he does not randomly select a door to open - he always opens a door with a goat. By doing this he reduced the possible scenarios for Jennifer to the four listed above. If he randomly picked, then Jenny's chances, if the show progressed as presented, would be 50/50. However, there would also be a 1:3 chance that MB would open the wrong door and reveal the car's location (followed by a 1:1 chance that MB would be sacked and re-runs of McGyver would fill out the remainder of the season!)
7. SEQUENCE PUZZLE
The below is a number puzzle. It should be read left to right, top to bottom.
Question 1: What is the next two rows of numbers?
Question 2: How was this reached?
1
1 1
2 1
1 2 1 1
1 1 1 2 2 1
? ? ? ? ? ?
? ? ? ? ? ? ? ?
ANS=
The secret is to say whay you see on each line and what you see becomes the next line. For full answer see below.
Line 1 is "Two ones" (2 1)
Line 2 then becomes "One two, and one one" (1 2 1 1)
Line 3 therefore is "One one, one two and two ones" (1 1 1 2 2 1)
Line 4 is "Three ones, two twos and one one" (3 1 2 2 1 1)
Line 5 is "One three, one one, two twos and two ones" (1 3 1 1 2 2 2 1)
Question 1: What is the next two rows of numbers?
Question 2: How was this reached?
1
1 1
2 1
1 2 1 1
1 1 1 2 2 1
? ? ? ? ? ?
? ? ? ? ? ? ? ?
ANS=
The secret is to say whay you see on each line and what you see becomes the next line. For full answer see below.
Line 1 is "Two ones" (2 1)
Line 2 then becomes "One two, and one one" (1 2 1 1)
Line 3 therefore is "One one, one two and two ones" (1 1 1 2 2 1)
Line 4 is "Three ones, two twos and one one" (3 1 2 2 1 1)
Line 5 is "One three, one one, two twos and two ones" (1 3 1 1 2 2 2 1)
Line 1 is "Two ones" (2 1)
Line 2 then becomes "One two, and one one" (1 2 1 1)
Line 3 therefore is "One one, one two and two ones" (1 1 1 2 2 1)
Line 4 is "Three ones, two twos and one one" (3 1 2 2 1 1)
Line 5 is "One three, one one, two twos and two ones" (1 3 1 1 2 2 2 1)
8. HOW DID IT HAPPEN?
A man runs a mile south, a mile west, and a mile north and ends up back where he started! How did it happen?
That might be easy but the harder part is that there are actually an infinite number of answers for where the man could have started from. Explain.
ANS=
Any point on the circle (1 + 1/2pi) miles from the South Pole.
After going South 1 mile, you're (1/2pi) miles from the Pole,
which allows you to run West 1 mile [1 lap of a 1-mile circumference circle]
and be able to go a mile North to the starting point.
There is an infinite number of starting distances:
1 + 1/2Npi miles North of the South pole where N is any positive integer.
N is then the number of circular laps in your westerly mile.
e.g. N=5280 - you'd run 5280 laps around a 1-foot circumference circle.
A man runs a mile south, a mile west, and a mile north and ends up back where he started! How did it happen?That might be easy but the harder part is that there are actually an infinite number of answers for where the man could have started from. Explain.
ANS=
Any point on the circle (1 + 1/2pi) miles from the South Pole.
After going South 1 mile, you're (1/2pi) miles from the Pole,
which allows you to run West 1 mile [1 lap of a 1-mile circumference circle]
and be able to go a mile North to the starting point.
There is an infinite number of starting distances:
1 + 1/2Npi miles North of the South pole where N is any positive integer.
N is then the number of circular laps in your westerly mile.
e.g. N=5280 - you'd run 5280 laps around a 1-foot circumference circle.
After going South 1 mile, you're (1/2pi) miles from the Pole,
which allows you to run West 1 mile [1 lap of a 1-mile circumference circle]
and be able to go a mile North to the starting point.
There is an infinite number of starting distances:
1 + 1/2Npi miles North of the South pole where N is any positive integer.
N is then the number of circular laps in your westerly mile.
e.g. N=5280 - you'd run 5280 laps around a 1-foot circumference circle.
9. HOLE IN A SPHERE
A 6-inch hole is drilled through a sphere. What is the volume of the remaining portion of the sphere?
Clarifications:
[1] the hole is a circular cylinder of empty space whose axis passes through the center of the sphere - just as a drill would make if you aimed the center of the drill at the center of the sphere and made sure you drilled all the way through.
[2] the length of the hole [6 inches] is the height of the cylinder that forms the inside surface once the hole is drilled. picture the inside surface as viewed from inside the hole and measure the length of that surface in the direction of the axis of the drill.
in this sense, you could for example drill a 6-inch hole through the earth. the diameter of the hole would be huge, and you'd just have a tiny remnant of the earth left. but if you could set it on a table [a big table] it would be 6 inches high.
You of course could not drill a 6-inch hole through a sphere whose diameter was less than 6 inches. This fact leads to the logical answer.
The hard way involves calculus. The easy way uses logic.
ANS=
Here's the mathematical solution:
Radius of sphere = R, radius of hole = r, length of hole = 2L [so L=3], height of cap = h.
Since r*r = R*R - L*L and h = R-L, we can eliminate r and h and do everything in terms of R and L.
Swizzling cpotting's cap formula, V[cap] = pi/3 (2R*R*R - 3R*R*L + L*L*L)
Cylinders are ho-hum, V[cyl] = 2pi*L*r*r = (2pi/3) (3R*R*L - 3L*L*L)
V[removed by drilling] = V[cylinder] + 2V[cap]
doing the math,
V[removed] = (4pi/3)R*R*R - (4pi/3)L*L*L
Pretty amazing: the volume removed by a hole of length 2L is the difference of the volumes of two spheres: one of radius R, the other of radius L.
So the remaining volume is simply the volume of a shpere with radius L. [hint-hint at the logical solution]
V[remainder] = (4pi/3)L*L*L = 36pi.
Here's the logical solution:
My friend wouldn't have posed a math problem [boring] and she wouldn't have left out critical information. Therefore the answer couldn't depend on the radius of the sphere. I chose a sphere size [radius=3] that would make 0 volume removed [a hole of length 6 and diameter 0]. With nothing removed, the remaining volume is the original volume: (4pi/3) 3*3*3.
So it's 36pi.
Clarifications:
[1] the hole is a circular cylinder of empty space whose axis passes through the center of the sphere - just as a drill would make if you aimed the center of the drill at the center of the sphere and made sure you drilled all the way through.
[2] the length of the hole [6 inches] is the height of the cylinder that forms the inside surface once the hole is drilled. picture the inside surface as viewed from inside the hole and measure the length of that surface in the direction of the axis of the drill.
in this sense, you could for example drill a 6-inch hole through the earth. the diameter of the hole would be huge, and you'd just have a tiny remnant of the earth left. but if you could set it on a table [a big table] it would be 6 inches high.
You of course could not drill a 6-inch hole through a sphere whose diameter was less than 6 inches. This fact leads to the logical answer.
The hard way involves calculus. The easy way uses logic.
ANS=
Here's the mathematical solution:
Radius of sphere = R, radius of hole = r, length of hole = 2L [so L=3], height of cap = h.
Since r*r = R*R - L*L and h = R-L, we can eliminate r and h and do everything in terms of R and L.
Swizzling cpotting's cap formula, V[cap] = pi/3 (2R*R*R - 3R*R*L + L*L*L)
Cylinders are ho-hum, V[cyl] = 2pi*L*r*r = (2pi/3) (3R*R*L - 3L*L*L)
V[removed by drilling] = V[cylinder] + 2V[cap]
doing the math,
V[removed] = (4pi/3)R*R*R - (4pi/3)L*L*L
Pretty amazing: the volume removed by a hole of length 2L is the difference of the volumes of two spheres: one of radius R, the other of radius L.
So the remaining volume is simply the volume of a shpere with radius L. [hint-hint at the logical solution]
V[remainder] = (4pi/3)L*L*L = 36pi.
Here's the logical solution:
My friend wouldn't have posed a math problem [boring] and she wouldn't have left out critical information. Therefore the answer couldn't depend on the radius of the sphere. I chose a sphere size [radius=3] that would make 0 volume removed [a hole of length 6 and diameter 0]. With nothing removed, the remaining volume is the original volume: (4pi/3) 3*3*3.
So it's 36pi.
Radius of sphere = R, radius of hole = r, length of hole = 2L [so L=3], height of cap = h.
Since r*r = R*R - L*L and h = R-L, we can eliminate r and h and do everything in terms of R and L.
Swizzling cpotting's cap formula, V[cap] = pi/3 (2R*R*R - 3R*R*L + L*L*L)
Cylinders are ho-hum, V[cyl] = 2pi*L*r*r = (2pi/3) (3R*R*L - 3L*L*L)
V[removed by drilling] = V[cylinder] + 2V[cap]
doing the math,
V[removed] = (4pi/3)R*R*R - (4pi/3)L*L*L
Pretty amazing: the volume removed by a hole of length 2L is the difference of the volumes of two spheres: one of radius R, the other of radius L.
So the remaining volume is simply the volume of a shpere with radius L. [hint-hint at the logical solution]
V[remainder] = (4pi/3)L*L*L = 36pi.
Here's the logical solution:
My friend wouldn't have posed a math problem [boring] and she wouldn't have left out critical information. Therefore the answer couldn't depend on the radius of the sphere. I chose a sphere size [radius=3] that would make 0 volume removed [a hole of length 6 and diameter 0]. With nothing removed, the remaining volume is the original volume: (4pi/3) 3*3*3.
So it's 36pi.
10. HELP! A REMAINDER IS CHASING ME
I just found a number with an interesting property:
When I divide it by 2, the remainder is 1.
When I divide it by 3, the remainder is 2.
When I divide it by 4, the remainder is 3.
When I divide it by 5, the remainder is 4.
When I divide it by 6, the remainder is 5.
When I divide it by 7, the remainder is 6.
When I divide it by 8, the remainder is 7.
When I divide it by 9, the remainder is 8.
When I divide it by 10, the remainder is 9.
It's not a small number, but it's not really big, either.
Find the smallest number with such property.
ANS=
2519
The number has to end in 9.
Looked brute force for small numbers.
59 and 119 were promising, but no cigar.
Then looked for agreement among
39 + multiples of 40,
69 + multiples of 70 and
89 + multiples of 90
Smallest one was 2519.
Still think of this as kind of brute force.
Maybe there is no elegant solution.
When I divide it by 2, the remainder is 1.
When I divide it by 3, the remainder is 2.
When I divide it by 4, the remainder is 3.
When I divide it by 5, the remainder is 4.
When I divide it by 6, the remainder is 5.
When I divide it by 7, the remainder is 6.
When I divide it by 8, the remainder is 7.
When I divide it by 9, the remainder is 8.
When I divide it by 10, the remainder is 9.
It's not a small number, but it's not really big, either.
Find the smallest number with such property.
ANS=
2519
The number has to end in 9.
Looked brute force for small numbers.
59 and 119 were promising, but no cigar.
Then looked for agreement among
39 + multiples of 40,
69 + multiples of 70 and
89 + multiples of 90
Smallest one was 2519.
Still think of this as kind of brute force.
Maybe there is no elegant solution.
The number has to end in 9.
Looked brute force for small numbers.
59 and 119 were promising, but no cigar.
Then looked for agreement among
39 + multiples of 40,
69 + multiples of 70 and
89 + multiples of 90
Smallest one was 2519.
Still think of this as kind of brute force.
Maybe there is no elegant solution.
BONUS! RIDDLE ENCORE
Sir, I bear a rhyme excelling
In mystic force and magic spelling
Celestial sprites elucidate
All my own striving can't relate
ANS=
It's Pi
3,1 4 1 5 9
2 6 5 3 5 8
9 7 9
3 2 3 8 4 6
Sir, I bear a rhyme excellingIn mystic force and magic spelling
Celestial sprites elucidate
All my own striving can't relate
ANS=
It's Pi
3,1 4 1 5 9
2 6 5 3 5 8
9 7 9
3 2 3 8 4 6
3,1 4 1 5 9
2 6 5 3 5 8
9 7 9
3 2 3 8 4 6
OLD FUNNY ONE
Right now Mum is 21 years older then her child. In 6 years her child will be 5 times younger than she.
Where is Daddy?
ANS=
The child is brand new in the womb, since .75 of a year is about 9 months. So the father is...well, ya know- pretty close by.
Where is Daddy?
ANS=
The child is brand new in the womb, since .75 of a year is about 9 months. So the father is...well, ya know- pretty close by.
TRAPPED
There is a room with no windows, doors, or any sort of opening, the walls are solid steel 10 feet thick, and you are trapped inside, left only with a saw and a table.
How do you escape?
ANS=
Fantasy logic problem where words have alternate meanings (usually related to homonyms)
You saw the table in half; but the two halves together; two halves make a whole (hole) and you crawl out through the hole.
How do you escape?
ANS=
Fantasy logic problem where words have alternate meanings (usually related to homonyms)
You saw the table in half; but the two halves together; two halves make a whole (hole) and you crawl out through the hole.
You saw the table in half; but the two halves together; two halves make a whole (hole) and you crawl out through the hole.
HOW MANY WERE GOING TO SAINT IVES?
As I was going to Saint Ives,
I crossed the path of seven wives.
Every wife had seven sacks,
Every sack had seven cats,
Every cat had seven kittens,
Kittens, cats, sacks, wives,
How many were going to Saint Ives?
ANS=
It is 1
You are PASSING those people. You are the ONLY person going from X to St Ives, they are going from St Ives to X
I crossed the path of seven wives.
Every wife had seven sacks,
Every sack had seven cats,
Every cat had seven kittens,
Kittens, cats, sacks, wives,
How many were going to Saint Ives?
ANS=
It is 1
You are PASSING those people. You are the ONLY person going from X to St Ives, they are going from St Ives to X
You are PASSING those people. You are the ONLY person going from X to St Ives, they are going from St Ives to X
ANOTHER OLDIE - CAN YOU GET THEM ALL?
- 26 L of the A
- 7 D of the W
- 7 W of the W
- 12 S of the Z
- 66 B of the B
- 52 C in a P (W J)
- 13 S in the U S F
- 18 H on a G C
- 39 B of the O T
- 5 T on a F
- 90 D in a R A
- 3 B M (S H T R)
- 32 is the T in D F at which W F
- 15 P in a R T
- 3 W on a T
- 100 C in a D
- 11 P in a F (S) T
- 12 M in a Y
- 13 is U F S
- 8 T on an O
- 29 D in F in a L Y
- 27 B in the N T
- 365 D in a Y
- 13 L in a B D
- 52 W in a Y
- 9 L of a C
- 60 M in an H
- 23 P of C in the H B
- 64 S on a C B
- 9 P in S A
- 6 B to an O in C
- 1000 Y in a M
- 15 M on a D M C
ANS=
1. 26 L of the A - 26 Letters of the Alphabet
2. 7 D of the W - 7 days of the Week
3. 7 W of the W - 7 wonders of the world
4. 12 S of the Z - 12 signs of the zodiac
5. 66 B of the B - 66 books of the bible
6. 52 C in a P (W J) - 52 cars in a pack (without jokers)
7. 13 S in the U S F - 13 stripes in the united states flag
8. 18 H on a G C - 18 holes on a golf course
9. 39 B of the O T - 39 books of the old testament
10. 5 T on a F - 5 toes on a foot
11. 90 D in a R A - 90 degrees in a right angel
12. 3 B M (S H T R) - 3 blind mice (see how they run)
13. 32 is the T in D F at which W F - 32 is the temperature in degrees ferenheight at which water freezes
14. 15 P in a R T - 15 players in a rugby team
15. 3 W on a T - 3 wheels on a tricycle
16. 100 C in a D - 100 cents in a dollar
17. 11 P in a F (S) T - 11 players in a football (soccer) team
18. 12 M in a Y - 12 months in a year
19. 13 is U F S - 13 is unlucky for some
20. 8 T on an O - 8 tenticles on an octopus
21. 29 D in F in a L Y - 29 days in Feburary in a leap year
22. 27 B in the N T - 27 books in the new testiment
23. 365 D in a Y - 365 days in a year
24. 13 L in a B D - 13 loaves in a baker's dozen
25. 52 W in a Y - 52 weeks in a year
26. 9 L of a C - 9 lives of a cat
27. 60 M in an H - 60 minutes in an hour
28. 23 P of C in the H B - 23 pairs of chromosomes in the human body
29. 64 S on a C B - 64 squares on checkers board
30. 9 P in S A - 9 provinces in south africa
31. 6 B to an O in C - 6 bowls to an over in cricket
32. 1000 Y in a M - 1000 years in a millenium
33. 15 M on a D M C - 15 men on a dead man's chest
LETTER PROBLEM
Enter the letters A, B, C, D, E once in each row and column (in the first picture just A, B, C, D). The clues outside the grid indicate which letter appears first from that direction. In the third puzzle, for example, D must be the leftmost letter in both the first and third rows, and C must be the bottommost letter in the first and second columns.
ANS=
- 26 L of the A
- 7 D of the W
- 7 W of the W
- 12 S of the Z
- 66 B of the B
- 52 C in a P (W J)
- 13 S in the U S F
- 18 H on a G C
- 39 B of the O T
- 5 T on a F
- 90 D in a R A
- 3 B M (S H T R)
- 32 is the T in D F at which W F
- 15 P in a R T
- 3 W on a T
- 100 C in a D
- 11 P in a F (S) T
- 12 M in a Y
- 13 is U F S
- 8 T on an O
- 29 D in F in a L Y
- 27 B in the N T
- 365 D in a Y
- 13 L in a B D
- 52 W in a Y
- 9 L of a C
- 60 M in an H
- 23 P of C in the H B
- 64 S on a C B
- 9 P in S A
- 6 B to an O in C
- 1000 Y in a M
- 15 M on a D M C
ANS=
1. 26 L of the A - 26 Letters of the Alphabet
2. 7 D of the W - 7 days of the Week
3. 7 W of the W - 7 wonders of the world
4. 12 S of the Z - 12 signs of the zodiac
5. 66 B of the B - 66 books of the bible
6. 52 C in a P (W J) - 52 cars in a pack (without jokers)
7. 13 S in the U S F - 13 stripes in the united states flag
8. 18 H on a G C - 18 holes on a golf course
9. 39 B of the O T - 39 books of the old testament
10. 5 T on a F - 5 toes on a foot
11. 90 D in a R A - 90 degrees in a right angel
12. 3 B M (S H T R) - 3 blind mice (see how they run)
13. 32 is the T in D F at which W F - 32 is the temperature in degrees ferenheight at which water freezes
14. 15 P in a R T - 15 players in a rugby team
15. 3 W on a T - 3 wheels on a tricycle
16. 100 C in a D - 100 cents in a dollar
17. 11 P in a F (S) T - 11 players in a football (soccer) team
18. 12 M in a Y - 12 months in a year
19. 13 is U F S - 13 is unlucky for some
20. 8 T on an O - 8 tenticles on an octopus
21. 29 D in F in a L Y - 29 days in Feburary in a leap year
22. 27 B in the N T - 27 books in the new testiment
23. 365 D in a Y - 365 days in a year
24. 13 L in a B D - 13 loaves in a baker's dozen
25. 52 W in a Y - 52 weeks in a year
26. 9 L of a C - 9 lives of a cat
27. 60 M in an H - 60 minutes in an hour
28. 23 P of C in the H B - 23 pairs of chromosomes in the human body
29. 64 S on a C B - 64 squares on checkers board
30. 9 P in S A - 9 provinces in south africa
31. 6 B to an O in C - 6 bowls to an over in cricket
32. 1000 Y in a M - 1000 years in a millenium
33. 15 M on a D M C - 15 men on a dead man's chest
2. 7 D of the W - 7 days of the Week
3. 7 W of the W - 7 wonders of the world
4. 12 S of the Z - 12 signs of the zodiac
5. 66 B of the B - 66 books of the bible
6. 52 C in a P (W J) - 52 cars in a pack (without jokers)
7. 13 S in the U S F - 13 stripes in the united states flag
8. 18 H on a G C - 18 holes on a golf course
9. 39 B of the O T - 39 books of the old testament
10. 5 T on a F - 5 toes on a foot
11. 90 D in a R A - 90 degrees in a right angel
12. 3 B M (S H T R) - 3 blind mice (see how they run)
13. 32 is the T in D F at which W F - 32 is the temperature in degrees ferenheight at which water freezes
14. 15 P in a R T - 15 players in a rugby team
15. 3 W on a T - 3 wheels on a tricycle
16. 100 C in a D - 100 cents in a dollar
17. 11 P in a F (S) T - 11 players in a football (soccer) team
18. 12 M in a Y - 12 months in a year
19. 13 is U F S - 13 is unlucky for some
20. 8 T on an O - 8 tenticles on an octopus
21. 29 D in F in a L Y - 29 days in Feburary in a leap year
22. 27 B in the N T - 27 books in the new testiment
23. 365 D in a Y - 365 days in a year
24. 13 L in a B D - 13 loaves in a baker's dozen
25. 52 W in a Y - 52 weeks in a year
26. 9 L of a C - 9 lives of a cat
27. 60 M in an H - 60 minutes in an hour
28. 23 P of C in the H B - 23 pairs of chromosomes in the human body
29. 64 S on a C B - 64 squares on checkers board
30. 9 P in S A - 9 provinces in south africa
31. 6 B to an O in C - 6 bowls to an over in cricket
32. 1000 Y in a M - 1000 years in a millenium
33. 15 M on a D M C - 15 men on a dead man's chest
LETTER PROBLEM
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DOMINO HUNT
There are 28 dominos in the picture (0-0, 0-1, 0-2, ..., 6-6). Unfortunately, the edges can not be distinguished from the middle lines and it is up to you to find out where the 28 domino pieces are placed.
Where are the edges?
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Where are the edges?
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CRISSCROSS
This is a number crossword puzzle. Enter one numeral character into each square. The clue consists of mathematical operations (e. g. "D.6 x 3" means, that the number you are looking for is three times bigger than number D.6).
Across:
A.1 = D.6 x 3
A.4 = A.7 x A.7
A.5 = (D.3 x 7) - 1
A.6 = A.1 + A.5
A.7 = D.3
Down:
D.1 = ?
D.2 = A.7 x 6
D.3 = D.6 + 6
D.4 = A.4 - 210
D.6 = (D.2 + 9) x 1/7
ANS=
Across:
A.1 = D.6 x 3
A.4 = A.7 x A.7
A.5 = (D.3 x 7) - 1
A.6 = A.1 + A.5
A.7 = D.3
Down:
D.1 = ?
D.2 = A.7 x 6
D.3 = D.6 + 6
D.4 = A.4 - 210
D.6 = (D.2 + 9) x 1/7
ANS=
NESSIE
A monster snake, 45 m long, hides under water. The picture below shows the area where the monster is hidden. Each square represents 1 meter. Parts of the snake's body are connected horizontally or vertically across the squares on the picture below. Above the surface, there is only his head (1), tail (45) and 23rd meter part of body (23). The yellow numbers on the margins show how many squares are occupied by the snake's body parts in the corresponding row or column. Black squares are stones, where the snake can't be.
So where is the snake?
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So where is the snake?
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HEXAGONS
Find a looped path through the diagram subject to the following constraints. The path proceeds from one hexagon to an adjacent hexagon through the center of each hexagon, passes through no hexagon more than once, does not go through any numbered hexagon, and never makes a acute angle turn (i.e., a turn at a 60o angle). Each number indicates how many of the adjacent hexagons are part of the path.
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MATCHSTICK EQUATIONS
Try to rectify a mistake by moving a single matchstick, to get the correct equation.
The following equation is made of 11 matches:
XI - V = IV (more solutions)
The following equation is made of 11 matches:
X + V = IV (more solutions)
The following equation is made of 10 matches:
L + L = L (more solutions)
The following equation is made of 12 matches:
VI = IV - III (more solutions)
The following equation is made of 14 matches:
XIV - V = XX
The following equation is made of 11 matches:
IX - IX = V
The following equation is made of 12 matches:
X = VIII - II
The following equation is made of 7 matches:
VII = I
ANS=
XI - V = IV (more solutions) ... X - VI = IV or XI - V = VI or XI - VI = V
X + V = IV (more solutions) ... IX - V = IV or X - VI = IV
L + L = L (more solutions) ... C - L = L or L + I = LI
VI = IV - III (more solutions) ... VI = IX - III or VI = IV + II
XIV - V = XX ... XV + V = XX
IX - IX = V ... IX - IV = V
X = VIII - II ... X - VIII = II
VII = I ... square root of 1
The following equation is made of 11 matches:
XI - V = IV (more solutions)
The following equation is made of 11 matches:
X + V = IV (more solutions)
The following equation is made of 10 matches:
L + L = L (more solutions)
The following equation is made of 12 matches:
VI = IV - III (more solutions)
The following equation is made of 14 matches:
XIV - V = XX
The following equation is made of 11 matches:
IX - IX = V
The following equation is made of 12 matches:
X = VIII - II
The following equation is made of 7 matches:
VII = I
ANS=
XI - V = IV (more solutions) ... X - VI = IV or XI - V = VI or XI - VI = V
X + V = IV (more solutions) ... IX - V = IV or X - VI = IV
L + L = L (more solutions) ... C - L = L or L + I = LI
VI = IV - III (more solutions) ... VI = IX - III or VI = IV + II
XIV - V = XX ... XV + V = XX
IX - IX = V ... IX - IV = V
X = VIII - II ... X - VIII = II
VII = I ... square root of 1
X + V = IV (more solutions) ... IX - V = IV or X - VI = IV
L + L = L (more solutions) ... C - L = L or L + I = LI
VI = IV - III (more solutions) ... VI = IX - III or VI = IV + II
XIV - V = XX ... XV + V = XX
IX - IX = V ... IX - IV = V
X = VIII - II ... X - VIII = II
VII = I ... square root of 1
TRICKY MATCHSTICK EQUATION
Move one matchstick to get the correct equation.
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Move one matchstick to get the correct equation.
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4 IDENTICAL TRIANGLES
Move one matchstick to get 4 identical triangles.
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Move one matchstick to get 4 identical triangles.
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SHOVEL
Move just two matches and remove dust from the shovel.
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ANS=
HOUSE
1. Move just two matches to make eleven squares.
2. Move four matches and form 15 squares.
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1. eleven squares
2. fifteen squares
2. Move four matches and form 15 squares.
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1. eleven squares
2. fifteen squares
2. fifteen squares
SCALES
Move 5 matches to make the scales balanced.
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.
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.FISH
Move just 3 matches so that the fish swims the other direction.
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RABBIT HUTCH
In the picture there are little flats for 6 rabbits. Can you build a dwelling for these 6 rabbits with only 12 matches? 6 separate flats are needed for the rabbits.
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COW
This cow has the following parts: head, body, horns, legs and tail. It is looking to the left. Move two matches so that it is looking to the right.
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KEY
1. Move four matches so that three squares are created.
2. Move three matches so that two rectangles are created.
3. Move two matches so that two rectangles are created.
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1. three squares
2. two rectangles
3. two rectangles
2. Move three matches so that two rectangles are created.
3. Move two matches so that two rectangles are created.
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1. three squares
2. two rectangles
3. two rectangles
2. two rectangles
3. two rectangles
TOUCH
Place six matches in such a way that each match is in touch with all the other five matches.
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Another alternative to arrange it is as follows.
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Another alternative to arrange it is as follows.
6X4 RECTANGLE
Your task is to dissect pictures to rearange the pieces into new shapes in the following brain puzzles.
Slice the picture into 2 sections from which you could make a rectangle 6x4 squares.
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Slice the picture into 2 sections from which you could make a rectangle 6x4 squares.
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8X8 SQUARE
Slice the picture into 2 sections from which you could make an 8x8 square.
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2 IDENTICAL SECTIONS
Slice the picture into 2 identical/symmetrical sections.
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CATERPILLARS
Slice the square into 4 identical sections, so that in each section there is 1 caterpillar with its leaf. One caterpillar will not have a leaf, she is taking a diet.
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LIFT
Slice the rectangle with a hole in its centre into 2 sections so that you could make a square 8x8 - virgin (without that hole in the centre).
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4 IDENTICAL SECTIONS
Slice the picture into 4 identical sections in this brain puzzle.
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FOLDING PAPER
The following brain puzzles are a bit different. No dissection is needed.
Divide a paper into 8 sections and write numbers on it according to the picture. Your job is to fold it where the lines are so that the numbers are sorted (number 1 will be on the top, 2 under it,..., and the last one will be 8).
Modification: It is the same objective, just the numbers have changed.
Divide a paper into 8 sections and write numbers on it according to the picture. Your job is to fold it where the lines are so that the numbers are sorted (number 1 will be on the top, 2 under it,..., and the last one will be 8).
Modification: It is the same objective, just the numbers have changed.
WRITE NUMBERS
Write the numbers from 1 to 8 into the squares, so that the squares with consecutive numbers do not touch (neither edges nor corners).
ANS=
ANS=
64 = 65 GEOMETRY PARADOX
Where does the hole in second triangle come from (the partitions are the same)?
ANS=
It looks like a triangle, because a thick line was used. Hypotenuse of the composite triangle is actually not a straight line - it is made of two lines. Forth cusps are where the arrows point (c9, l6).
The 64 = 65 paradox arises from the fact that the edges of the four pieces, which lie along the diagonal of the formed rectangle, do not coincide exactly in direction. This diagonal is not a straight segment line but a small lozenge (diamond-shaped figure), whose acute angle is
arctan 2/3 - arctan 3/8 = arctan 1/46
which is less than 1 degree 15' . Only a very precise drawing can enable us to distinguish such a small angle. Using analytic geometry or trigonometry, we can easily prove that the area of the "hidden" lozenge is equal to that of a small square of the chessboard.
ANS=
It looks like a triangle, because a thick line was used. Hypotenuse of the composite triangle is actually not a straight line - it is made of two lines. Forth cusps are where the arrows point (c9, l6).
The 64 = 65 paradox arises from the fact that the edges of the four pieces, which lie along the diagonal of the formed rectangle, do not coincide exactly in direction. This diagonal is not a straight segment line but a small lozenge (diamond-shaped figure), whose acute angle is
arctan 2/3 - arctan 3/8 = arctan 1/46
which is less than 1 degree 15' . Only a very precise drawing can enable us to distinguish such a small angle. Using analytic geometry or trigonometry, we can easily prove that the area of the "hidden" lozenge is equal to that of a small square of the chessboard.
The 64 = 65 paradox arises from the fact that the edges of the four pieces, which lie along the diagonal of the formed rectangle, do not coincide exactly in direction. This diagonal is not a straight segment line but a small lozenge (diamond-shaped figure), whose acute angle is
arctan 2/3 - arctan 3/8 = arctan 1/46
which is less than 1 degree 15' . Only a very precise drawing can enable us to distinguish such a small angle. Using analytic geometry or trigonometry, we can easily prove that the area of the "hidden" lozenge is equal to that of a small square of the chessboard.
GEOMETRY PARADOX - ANOTHER VARIATION
The same principle - moving the same parts - allows creating objects 64, 65 and 63 squares big. This geometric fallacy is also known as '64 = 65 Geometry Paradox'.
ANS=
Check explanation of the previous geometry paradox. The same principle applies here as well.
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Check explanation of the previous geometry paradox. The same principle applies here as well.
FIVE VIEWS
The five images below represent five views (from five of its six sides) of a solid object. This object has been assembled by gluing together several identical small cubes so that at least one face of each small cube is totally adherent to a face of another contiguous small cube. Each black line shown represents a side of the object that is perpendicular to the plane of this page. Draw the sixth view of the object and calculate how many small cubes were used in the construction of the object, as well as its respective colors.
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HOEFLIN'S OBJECT I
Several identical cubes are fused together to form a solid object. Given the following five external views of such an object, draw the sixth external view. Clockwise or counterclockwise rotations of the sixth view are acceptable, but a mirror image (the sixth side as viewed from inside the solid) is not acceptable.
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HOEFLIN'S OBJECT II
The five figures shown below represent the appearance of a solid, opaque object as seen from five of its six sides. Each line shown depicts a side of the object that is perpendicular to the plane of this page. The object was constructed by gluing together a number of identical cubes so that at least one face of each added cube precisely and entirely covers and is everywhere contiguous with one face of a previous cube. Draw the sixth view of the object.
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The five figures shown below represent the appearance of a solid, opaque object as seen from five of its six sides. Each line shown depicts a side of the object that is perpendicular to the plane of this page. The object was constructed by gluing together a number of identical cubes so that at least one face of each added cube precisely and entirely covers and is everywhere contiguous with one face of a previous cube. Draw the sixth view of the object.
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NUMBER SEQUENCES
There are infinite formulas that will fit any finite series. Try to guess the following number in each sequence (using the most simple mathematical operations, because as I mentioned, there is more than one solution for each number sequence).
- 8723, 3872, 2387, ?
- 1, 4, 9, 18, 35, ?
- 23, 45, 89, 177, ?
- 7, 5, 8, 4, 9, 3, ?
- 11, 19, 14, 22, 17, 25, ?
- 3, 8, 15, 24, 35, ?
- 2, 4, 5, 10, 12, 24, 27, ?
- 1, 3, 4, 7, 11, 18, ?
- 99, 92, 86, 81, 77, ?
- 0, 4, 2, 6, 4, 8, ?
- 1, 2, 2, 4, 8, 11, 33, ?
- 1, 2, 6, 24, 120, ?
- 1, 2, 3, 6, 11, 20, 37, ?
- 5, 7, 12, 19, 31, 50, ?
- 27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, ?
- 126, 63, 190, 95, 286, 143, 430, 215, 646, 323, 970, ?
- 4, 7, 15, 29, 59, 117, ?
- 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, ?
- 4, 4, 341, 6, 4, 4, 6, 6, 4, 4, 6, 10, 4, 4, 14, 6, 4, 4, 6, 6, 4, 4, 6, 22, 4, 4, 9, 6, ?
ANS=
- 8723, 3872, 2387, ? 7238 (moving of numerals)
- 1, 4, 9, 18, 35, ? 68 (x*2+2, +1, +0, -1, -2)
- 23, 45, 89, 177, ? 353 (x*2-1)
- 7, 5, 8, 4, 9, 3, ? 10, 2 (two series - every second number: 7, 8, 9, 10 and 5, 4, 3, 2)
- 11, 19, 14, 22, 17, 25, ? 20, 28 (two series - every second number: 11, 14, 17, 20 and 19, 22, 25, 28)
- 3, 8, 15, 24, 35, ? 48 (x+5, +7, +9, +11, +13)
- 2, 4, 5, 10, 12, 24, 27, ? 54, 58 (x*2, +1, *2, +2, *2, +3, *2, +4)
- 1, 3, 4, 7, 11, 18, ? 29 (a+b=c, b+c=d, c+d=e)
- 99, 92, 86, 81, 77, ? 74 (x-7, -6, -5, -4, -3)
- 0, 4, 2, 6, 4, 8, ? 6 (x+4, -2, +4, -2, +4, -2)
- 1, 2, 2, 4, 8, 11, 33, ?
- 1, 2, 6, 24, 120, ? 720 (x*2, *3, *4, *5, *6)
- 1, 2, 3, 6, 11, 20, 37, ?
- 5, 7, 12, 19, 31, 50, ? 81 (a+b=c, b+c=d, c+d=e)
- 27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, ? 322, 161 (x*3+1, /2, *3+1, /2)
- 126, 63, 190, 95, 286, 143, 430, 215, 646, 323, 970, ? 485, 1456 (x/2, *3+1, /2, *3+1)
- 4, 7, 15, 29, 59, 117, ? 235 (x*2-1, *2+1, *2-1)
- 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, ? 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5
- 4, 4, 341, 6, 4, 4, 6, 6, 4, 4, 6, 10, 4, 4, 14, 6, 4, 4, 6, 6, 4, 4, 6, 22, 4, 4, 9, 6, ? 4, 4
There are infinite formulas that will fit any finite series. Try to guess the following number in each sequence (using the most simple mathematical operations, because as I mentioned, there is more than one solution for each number sequence).
- 8723, 3872, 2387, ?
- 1, 4, 9, 18, 35, ?
- 23, 45, 89, 177, ?
- 7, 5, 8, 4, 9, 3, ?
- 11, 19, 14, 22, 17, 25, ?
- 3, 8, 15, 24, 35, ?
- 2, 4, 5, 10, 12, 24, 27, ?
- 1, 3, 4, 7, 11, 18, ?
- 99, 92, 86, 81, 77, ?
- 0, 4, 2, 6, 4, 8, ?
- 1, 2, 2, 4, 8, 11, 33, ?
- 1, 2, 6, 24, 120, ?
- 1, 2, 3, 6, 11, 20, 37, ?
- 5, 7, 12, 19, 31, 50, ?
- 27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, ?
- 126, 63, 190, 95, 286, 143, 430, 215, 646, 323, 970, ?
- 4, 7, 15, 29, 59, 117, ?
- 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, ?
- 4, 4, 341, 6, 4, 4, 6, 6, 4, 4, 6, 10, 4, 4, 14, 6, 4, 4, 6, 6, 4, 4, 6, 22, 4, 4, 9, 6, ?
ANS=
- 8723, 3872, 2387, ? 7238 (moving of numerals)
- 1, 4, 9, 18, 35, ? 68 (x*2+2, +1, +0, -1, -2)
- 23, 45, 89, 177, ? 353 (x*2-1)
- 7, 5, 8, 4, 9, 3, ? 10, 2 (two series - every second number: 7, 8, 9, 10 and 5, 4, 3, 2)
- 11, 19, 14, 22, 17, 25, ? 20, 28 (two series - every second number: 11, 14, 17, 20 and 19, 22, 25, 28)
- 3, 8, 15, 24, 35, ? 48 (x+5, +7, +9, +11, +13)
- 2, 4, 5, 10, 12, 24, 27, ? 54, 58 (x*2, +1, *2, +2, *2, +3, *2, +4)
- 1, 3, 4, 7, 11, 18, ? 29 (a+b=c, b+c=d, c+d=e)
- 99, 92, 86, 81, 77, ? 74 (x-7, -6, -5, -4, -3)
- 0, 4, 2, 6, 4, 8, ? 6 (x+4, -2, +4, -2, +4, -2)
- 1, 2, 2, 4, 8, 11, 33, ?
- 1, 2, 6, 24, 120, ? 720 (x*2, *3, *4, *5, *6)
- 1, 2, 3, 6, 11, 20, 37, ?
- 5, 7, 12, 19, 31, 50, ? 81 (a+b=c, b+c=d, c+d=e)
- 27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, ? 322, 161 (x*3+1, /2, *3+1, /2)
- 126, 63, 190, 95, 286, 143, 430, 215, 646, 323, 970, ? 485, 1456 (x/2, *3+1, /2, *3+1)
- 4, 7, 15, 29, 59, 117, ? 235 (x*2-1, *2+1, *2-1)
- 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, ? 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5
- 4, 4, 341, 6, 4, 4, 6, 6, 4, 4, 6, 10, 4, 4, 14, 6, 4, 4, 6, 6, 4, 4, 6, 22, 4, 4, 9, 6, ? 4, 4
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