It is likely that the found, sometimes quite cunning solutions will be useful to you in real life.
1. Cheryl’s birthday
Suppose some Bernard and Albert recently met a girl Cheryl. They want to find out when she has a birthday – to prepare gifts. But shelil is still a little thing. Instead of answering, she gives the guys a list of 10 possible dates:
|
May 15 |
16th of May |
May 19 |
|
June 17 |
June 18 |
|
|
the 14 th of July |
July 16 |
|
|
August 14 |
August 15 |
August 17 |
Predictably discovering that young men cannot calculate the correct date, Cheryl in a whisper, in his ear, calls Albert only a month of her birth. And Bernard is also quiet – only the number.
“Hmm,” says Albert. – I don’t know when Cheryl’s birthday. But I know for sure that Bernard does not know this either.
“Ha,” Bernard answers. – At first I also did not know when Cheryl’s birthday, but now I know this!
“Yeah,” Albert agrees. – Now I know too.
And they call the right date in unison. When is Cheryl’s birthday?
If you can’t find an answer on the move, don’t be discouraged. For the first time, this question was born at a teenage mathematical Olympiad in Singapore, which is famous for the highest educational standards. After one of the local TV presenters published a screen of this problem on Facebook*, it became viral: tens of thousands of users Facebook*, Twitter, Reddit tried to solve it. But not everyone coped.
We are sure that you will succeed. Do not open a guess until at least you try.
July 16. This follows from the dialogue between Albert and Bernard. Plus a little exception method. Look.
If Cheryl was born in May or June, then her birthday can be 19th or 18th or 18th. These numbers are found in the list only once. Accordingly, Bernard, having heard them, could immediately understand what month we are talking about. But Albert, as follows from his first remark, is sure that Bernard, knowing the number, will definitely not be able to name the month. So, this is not about May or June. Cheryl was born in a month, each of the named dates in which has a double in the neighboring months. That is – in July or August.
Bernard, who knows the number of births by hearing and analyzing Albert’s replica (that is, finding out about July or August), reports that he now knows the correct answer. It follows that the number known to Bernard is not 14, because it is duplicated both in July and in August, so you cannot determine the right date. But Bernard is sure of his decision. So, the number he knew has no takes in July and August. Three options fall under this condition: July 16, August 15 and August 17.
In turn, Albert, having heard the words of Bernard (and logically reaching three of the above possible dates), states that now he also knows the right date. We remember that Albert is known for the month. If this month was August, the young man could not determine the number – after all, in August two appear at once. So, there is only one possible option – July 16.
View answer
2. How many years are daughters
Two former classmates once met on the street, and such a dialogue took place between them.
– Hello!
– Hello!
– How are you?
– Fine. Two daughters are growing, preschool children still.
– And how old are they?
– Well -oo -oo … the work of their ages is equal to the number of pigeons under our feet.
– This information is not enough for me!
– The eldest looks like a mother.
– Now I know the answer to my question!
So how old are the daughters of one of the interlocutors?
1 and 4 years. Since the answer became clear only after receiving information that one of the daughters is older, which means that there was an ambiguity before. At first, based on the number of pigeons, the option was considered that the daughter is twins (that is, their ages are equal). This is possible only with the number of pigeons equal to the squares of numbers up to 7 inclusive (7 years – the age when children go to school, that is, stop being preschoolers): 1, 4, 9, 16, 25, 36, 49.
Of these squares, only one can be obtained by multiplying two different numbers, each of which is equal or less than 7, 4 (1 × 4). Respectively, daughters 1 and 4 years. There are no other whole and at the same time “preschool” options.
View answer
3. Where is my car?!
They say that this task is set for primary school students Hong Kong Schools. Children solve it literally in a matter of seconds.
What is the number of the parking space that occupies the car.
87. To guess, just look at the picture from the other side. Then the numbers that you now see upside down will occupy the correct position – 86, 87, 88, 89, 90, 91.
View answer
4. Love in kleptopia
Jan and Maria fell in love with each other, communicating only through the Internet. Jan wants to send Maria a wedding ring by mail – to make an offer. But the trouble is: lovers live in the country of kleptopia, where any parcel transmitted by mail will certainly be stolen – unless it is enclosed in a box with a lock.
Jan and Mary have many castles, but they cannot send the keys to each other – because the keys will also be stolen. How to send a ring to Yana so that it probably gets to Mary in the hands?
Jan must send Maria a ring in a box locked on the castle. Without a key, naturally. Maria, having received the parcel, should cut her own castle into it.
Then the box goes to Jan again. He opens his castle with his own key and again addresses the package with the only remaining locked castle of Mary. And the girl has a key to him.
By the way, this task is not just a theoretical game for logic. The idea used in it is fundamental in the cryptographic principle of the exchange of keys to the protocol of Diffy – Hellman. This protocol allows two or more sides to get a common secret key using a communication channel unprotected from listening.
View answer
5. In search of fakes
The courier brought you 10 bags, in each of which there are a lot of coins. And everything would be fine, but you suspect that the money in one of the bags is false. All that you know for sure is that real coins weigh 1 g each, and fake – 1.1 g. There are no other differences between money.
Fortunately, you have accurate digital scales showing weight up to the tenth gram. But the courier is in a hurry.
In a word, there is no time, you are given only one attempt to use the scales. How to accurately calculate for one weighing, in which bag there are fake coins and whether there is such a bag at all?
One weighing is enough. Just put 55 coins on the scales at once: 1 – from the first bag, 2 – from the second, 3 – from the third, 4 – from the fourth … 10 – from the tenth. If the whole bunch of money will weigh 55 g, then there are no fake in any of the bags. But if the weight is different, you will immediately understand what is the serial number of the bag full of fakes.
Consider: if the readings of the scales differ from the reference by 0.1 – fake coins in the first bag, by 0.2 – in the second, by 0.3 – in the third … by 1.0 – in the tenth of the tenth.
View answer
6. The equality of tanks
In the dark -dark one (you can’t see ZGI in general, and you can’t turn on the light) the room is worth the table on which there are 50 coins. You do not see them, but you can feel them, turn over. And most importantly, you know for sure: 40 coins are originally lying upwards, and 10 – by the river.
Your task is to divide the money into two groups (not necessarily equal), each of which will have the same amount.
Divide coins into two groups: in one 40, in the other 10. Now turn all the money from the second group turn over all the money. Voila, you can include light: the task is completed. Do not believe – check.
For letter -eaters, we explain the algorithm. After the blind division into two groups, this happened: in the first there was a x nurse;And in the second, respectively, – (10 – x) naps (after all, in total according to the conditions of the problem of salescaps 10). And Orlov, thus, is 10 – (10 – x) = x. That is, the number of eagles in the second group is equal to the number of sales in the first.
We take the simplest step – we turn all the coins in the second heap. Thus, all the oral coins (x pieces) become coins -decisions, and their number turns out to be the same as the number of cuts in the first group.
View answer
7. How to not get married
Once the owner of a small shop in Italy owed a large sum to the usurer. He did not have the opportunity to pay off the debt. But there was a beauty, who had long liked the lender.
“Let’s do it,” the Rostovman suggested the shopkeeper. – You will give my daughter for me, and I forget about the duty to the same. Well, hand?
But the girl did not want to marry an old and ugly man. Therefore, the shopkeeper refused. However, potential son -in -law caught in his voice the hesitation and made a new proposal.
“I don’t want to force anyone,” said the user. – Let the case decide everything for us. See: I will put two stones in a bag – black and white. And let the daughter not look out one of them. If it is black, we will get married and I will forgive you a duty. If white – I will forgive the debt just like that, without demanding the hand of your daughter.
The deal looked fair, and this time my father agreed. The moneylender leaned toward the path, strewn with pebbles, quickly raised the stones and put them in a bag. But the daughter noticed terrible: both pebbles were black! Whatever she pulled out, she would have to get married. Of course, it was possible to convict a trap in deception, taking out both stones at once. But he could be furious and cancel the transaction, having requested the debt in full.
Thinking a couple of seconds, the girl confidently held out her hand to the bag. And she did something that saved her father from duty, and her herself – from the need for marriage. The justice of her act was recognized even by a usurer. What exactly she did?
The girl pulled out a stone and, not having time to show anyone, as if accidentally dropped it on the path. The pebble immediately mixed with the rest.
– Oh, I’m so awkward! – the daughter of the shopkeeper threw up her hands. – But it’s nothing. After all, we can look into the bag. If there is a white stone remained there, then I pulled out black. And vice versa.
Of course, when everyone looked into the bag, a black stone was discovered there. Even the moneylender was forced to agree: this means that the girl pulled out white. And if so – there will be no wedding and there will be a debt to forgive.
View answer
8. Your code is confused ..
You closed the suitcase on a digital lock with a three -digit code and accidentally forgot the numbers. But the memory offers you the following tips:
- 682 – in this code one of the numbers is true and stands in its place;
- 614 – one of the numbers is correct, but it is not in its place;
- 206 – two digits are true, but both are not in their places;
- 738 – generally nonsense, not a single hit;
- 870 – one figure is true, but not in its place.
This information is enough to choose the right code. What he is?
042.
Following the fourth clue, we cross out of all combinations the numbers 7, 3 and 8 – there are definitely no them in the sophisticated code. From the first clue, we find out that either 6 or 2 takes its place. But if this is 6, then the condition of the second clue is not fulfilled, where 6 stands at the beginning. So, the last number of code is 2. And 6 in the cipher is completely absent.
From the third hint, we conclude that the correct code numbers are 2 and 0. At the same time, 2 is in last place. So, 0 – on the first. Thus, the first and third digits of the code become known to us: 0 … 2.
We check with the second hint. The number 6 shallow earlier. The unit is not suitable: it is known that it is not in its place, but all possible places for it – the first and last – are already busy. Thus, only the number 4 is true. It and move it in the middle of the resulting code – 042.
View answer
9. How to divide the cake
And finally a little sweet. You have a festive cake, which must be divided by the number of guests – into 8 pieces. The only problem is that this must be done by completing only three cuts. Cope?
Make two cuts of the cross -steam cross – as if you want to divide the cake into four equal parts. And draw the third incision not vertically, but horizontally, dividing the treat along.
Fabiosa.Make a look at the answer
1. Cheryl’s birthday
Suppose some Bernard and Albert recently met a girl Cheryl. They want to find out when she has a birthday – to prepare gifts. But shelil is still a little thing. Instead of answering, she gives the guys a list of 10 possible dates:
|
May 15 |
16th of May |
May 19 |
|
June 17 |
June 18 |
|
|
the 14 th of July |
July 16 |
|
|
August 14 |
August 15 |
August 17 |
Predictably discovering that young men cannot calculate the correct date, Cheryl in a whisper, in his ear, calls Albert only a month of her birth. And Bernard is also quiet – only the number.
“Hmm,” says Albert. – I don’t know when Cheryl’s birthday. But I know for sure that Bernard does not know this either.
“Ha,” Bernard answers. – At first I also did not know when Cheryl’s birthday, but now I know this!
“Yeah,” Albert agrees. – Now I know too.
And they call the right date in unison. When is Cheryl’s birthday?
If you can’t find an answer on the move, don’t be discouraged. For the first time, this question was born at a teenage mathematical Olympiad in Singapore, which is famous for the highest educational standards. After one of the local TV presenters published a screen of this problem on Facebook*, it became viral: tens of thousands of users Facebook*, Twitter, Reddit tried to solve it. But not everyone coped.
We are sure that you will succeed. Do not open a guess until at least you try.
July 16. This follows from the dialogue between Albert and Bernard. Plus a little exception method. Look.
If Cheryl was born in May or June, then her birthday can be 19th or 18th or 18th. These numbers are found in the list only once. Accordingly, Bernard, having heard them, could immediately understand what month we are talking about. But Albert, as follows from his first remark, is sure that Bernard, knowing the number, will definitely not be able to name the month. So, this is not about May or June. Cheryl was born in a month, each of the named dates in which has a double in the neighboring months. That is – in July or August.
Bernard, who knows the number of births by hearing and analyzing Albert’s replica (that is, finding out about July or August), reports that he now knows the correct answer. It follows that the number known to Bernard is not 14, because it is duplicated both in July and in August, so you cannot determine the right date. But Bernard is sure of his decision. So, the number he knew has no takes in July and August. Three options fall under this condition: July 16, August 15 and August 17.
In turn, Albert, having heard the words of Bernard (and logically reaching three of the above possible dates), states that now he also knows the right date. We remember that Albert is known for the month. If this month was August, the young man could not determine the number – after all, in August two appear at once. So, there is only one possible option – July 16.
View answer
2. How many years are daughters
Two former classmates once met on the street, and such a dialogue took place between them.
– Hello!
– Hello!
– How are you?
– Fine. Two daughters are growing, preschool children still.
– And how old are they?
– Well -oo -oo … the work of their ages is equal to the number of pigeons under our feet.
– This information is not enough for me!
– The eldest looks like a mother.
– Now I know the answer to my question!
So how old are the daughters of one of the interlocutors?
1 and 4 years. Since the answer became clear only after receiving information that one of the daughters is older, which means that there was an ambiguity before. At first, based on the number of pigeons, the option was considered that the daughter is twins (that is, their ages are equal). This is possible only with the number of pigeons equal to the squares of numbers up to 7 inclusive (7 years – the age when children go to school, that is, stop being preschoolers): 1, 4, 9, 16, 25, 36, 49.
Of these squares, only one can be obtained by multiplying two different numbers, each of which is equal or less than 7, 4 (1 × 4). Respectively, daughters 1 and 4 years. There are no other whole and at the same time “preschool” options.
View answer
3. Where is my car?!
They say that this task is set for primary school students Hong Kong Schools. Children solve it literally in a matter of seconds.
What is the number of the parking space that occupies the car.
87. To guess, just look at the picture from the other side. Then the numbers that you now see upside down will occupy the correct position – 86, 87, 88, 89, 90, 91.
View answer
4. Love in kleptopia
Jan and Maria fell in love with each other, communicating only through the Internet. Jan wants to send Maria a wedding ring by mail – to make an offer. But the trouble is: lovers live in the country of kleptopia, where any parcel transmitted by mail will certainly be stolen – unless it is enclosed in a box with a lock.
Jan and Mary have many castles, but they cannot send the keys to each other – because the keys will also be stolen. How to send a ring to Yana so that it probably gets to Mary in the hands?
Jan must send Maria a ring in a box locked on the castle. Without a key, naturally. Maria, having received the parcel, should cut her own castle into it.
Then the box goes to Jan again. He opens his castle with his own key and again addresses the package with the only remaining locked castle of Mary. And the girl has a key to him.
By the way, this task is not just a theoretical game for logic. The idea used in it is fundamental in the cryptographic principle of the exchange of keys to the protocol of Diffy – Hellman. This protocol allows two or more sides to get a common secret key using a communication channel unprotected from listening.
View answer
5. In search of fakes
The courier brought you 10 bags, in each of which there are a lot of coins. And everything would be fine, but you suspect that the money in one of the bags is false. All that you know for sure is that real coins weigh 1 g each, and fake – 1.1 g. There are no other differences between money.
Fortunately, you have accurate digital scales showing weight up to the tenth gram. But the courier is in a hurry.
In a word, there is no time, you are given only one attempt to use the scales. How to accurately calculate for one weighing, in which bag there are fake coins and whether there is such a bag at all?
One weighing is enough. Just put 55 coins on the scales at once: 1 – from the first bag, 2 – from the second, 3 – from the third, 4 – from the fourth … 10 – from the tenth. If the whole bunch of money will weigh 55 g, then there are no fake in any of the bags. But if the weight is different, you will immediately understand what is the serial number of the bag full of fakes.
Consider: if the readings of the scales differ from the reference by 0.1 – fake coins in the first bag, by 0.2 – in the second, by 0.3 – in the third … by 1.0 – in the tenth of the tenth.
View answer
6. The equality of tanks
In the dark -dark one (you can’t see ZGI in general, and you can’t turn on the light) the room is worth the table on which there are 50 coins. You do not see them, but you can feel them, turn over. And most importantly, you know for sure: 40 coins are originally lying upwards, and 10 – by the river.
Your task is to divide the money into two groups (not necessarily equal), each of which will have the same amount.
Divide coins into two groups: in one 40, in the other 10. Now turn all the money from the second group turn over all the money. Voila, you can include light: the task is completed. Do not believe – check.
For letter -eaters, we explain the algorithm. After the blind division into two groups, this happened: in the first there was a x nurse;And in the second, respectively, – (10 – x) naps (after all, in total according to the conditions of the problem of salescaps 10). And Orlov, thus, is 10 – (10 – x) = x. That is, the number of eagles in the second group is equal to the number of sales in the first.
We take the simplest step – we turn all the coins in the second heap. Thus, all the oral coins (x pieces) become coins -decisions, and their number turns out to be the same as the number of cuts in the first group.
View answer
7. How to not get married
Once the owner of a small shop in Italy owed a large sum to the usurer. He did not have the opportunity to pay off the debt. But there was a beauty, who had long liked the lender.
“Let’s do it,” the Rostovman suggested the shopkeeper. – You will give my daughter for me, and I forget about the duty to the same. Well, hand?
But the girl did not want to marry an old and ugly man. Therefore, the shopkeeper refused. However, potential son -in -law caught in his voice the hesitation and made a new proposal.
“I don’t want to force anyone,” said the user. – Let the case decide everything for us. See: I will put two stones in a bag – black and white. And let the daughter not look out one of them. If it is black, we will get married and I will forgive you a duty. If white – I will forgive the debt just like that, without demanding the hand of your daughter.
The deal looked fair, and this time my father agreed. The moneylender leaned toward the path, strewn with pebbles, quickly raised the stones and put them in a bag. But the daughter noticed terrible: both pebbles were black! Whatever she pulled out, she would have to get married. Of course, it was possible to convict a trap in deception, taking out both stones at once. But he could be furious and cancel the transaction, having requested the debt in full.
Thinking a couple of seconds, the girl confidently held out her hand to the bag. And she did something that saved her father from duty, and her herself – from the need for marriage. The justice of her act was recognized even by a usurer. What exactly she did?
The girl pulled out a stone and, not having time to show anyone, as if accidentally dropped it on the path. The pebble immediately mixed with the rest.
– Oh, I’m so awkward! – the daughter of the shopkeeper threw up her hands. – But it’s nothing. After all, we can look into the bag. If there is a white stone remained there, then I pulled out black. And vice versa.
Of course, when everyone looked into the bag, a black stone was discovered there. Even the moneylender was forced to agree: this means that the girl pulled out white. And if so – there will be no wedding and there will be a debt to forgive.
View answer
8. Your code is confused ..
You closed the suitcase on a digital lock with a three -digit code and accidentally forgot the numbers. But the memory offers you the following tips:
- 682 – in this code one of the numbers is true and stands in its place;
- 614 – one of the numbers is correct, but it is not in its place;
- 206 – two digits are true, but both are not in their places;
- 738 – generally nonsense, not a single hit;
- 870 – one figure is true, but not in its place.
This information is enough to choose the right code. What he is?
042.
Following the fourth clue, we cross out of all combinations the numbers 7, 3 and 8 – there are definitely no them in the sophisticated code. From the first clue, we find out that either 6 or 2 takes its place. But if this is 6, then the condition of the second clue is not fulfilled, where 6 stands at the beginning. So, the last number of code is 2. And 6 in the cipher is completely absent.
From the third hint, we conclude that the correct code numbers are 2 and 0. At the same time, 2 is in last place. So, 0 – on the first. Thus, the first and third digits of the code become known to us: 0 … 2.
We check with the second hint. The number 6 shallow earlier. The unit is not suitable: it is known that it is not in its place, but all possible places for it – the first and last – are already busy. Thus, only the number 4 is true. It and move it in the middle of the resulting code – 042.
View answer
9. How to divide the cake
And finally a little sweet. You have a festive cake, which must be divided by the number of guests – into 8 pieces. The only problem is that this must be done by completing only three cuts. Cope?
Make two cuts of the cross -steam cross – as if you want to divide the cake into four equal parts. And draw the third incision not vertically, but horizontally, dividing the treat along.
Fabiosa.Make a look at the answer