9 logical problems that only intellectuals can handle

1. Cheryl's birthday

Suppose a certain Bernard and Albert recently met Cheryl's girlfriend. They want to know when her birthday is so they can prepare gifts. But Cheryl is such a thing. Instead of answering, she hands 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	17 august

Predictably, discovering that the young men cannot calculate the correct date, Cheryl, in a whisper in her ear, calls Alberta only the month of her birth. And Bernard - just as quiet - just a number.

“Hmm,” Albert says. “I don’t know when Cheryl’s birthday is.” But I know for a fact that Bernard doesn't know that either.

“Ha,” Bernard says. - At first I also did not know when Cheryl's birthday, but now I know it!

“Yeah,” Albert agrees. “Now I know too.

And they name the correct date in chorus. When is Cheryl's birthday?

If you can't find the answer right off the bat, don't be discouraged. This question was first asked at the Singapore and Asian School Math Olympiad, which is renowned for the highest educational standards in Singapore. After one of the local TV presenters posted a screen of this problem on Facebook, it went viral When is Cheryl’s birthday?’The tricky maths problem that has everyone stumped: tens of thousands of Facebook, Twitter, Reddit users tried to solve it. But not everyone did it.

We are confident that you will succeed. Don't open the answer until you at least try it.

July 16. This follows from the dialogue that took place between Albert and Bernard. Plus a little bit of an exception method. Look.

If Cheryl was born in May or June, then her birthday might be 19th or 18th. These numbers appear only once in the list. Accordingly, Bernard, hearing them, would immediately be able to understand what month they were talking about. But Albert, as follows from his first remark, is sure that Bernard, knowing the date, will definitely not be able to name the month. This means that we are not talking about May or June. Cheryl was born in a month, each of the named dates in which has a double in adjacent months. That is, in July or August.

Bernard, who knows the birth number, after hearing and analyzing Albert's remark (that is, finding out about July or August), reports that he now knows the correct answer. It follows from this that the number known to Bernard is not 14, because it is duplicated in July and August, so it is impossible to determine the correct date. But Bernard is confident in his decision. This means that the number known to him does not have duplicates in July and August. Three options fall under this condition: July 16, August 15 and August 17.

In turn, Albert, having heard Bernard's words (and logically reaching the three above-mentioned possible dates), declares that now he also knows the correct date. We remember that Albert knows the month. If this month were August, the young man would not have been able to determine the number - after all, in August there are two at once. This means that there is only one possible option - July 16.

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2. How old are the daughters

On the street, two former classmates once met, and such a dialogue took place between them.

- Hey!

- How are you?

- Good. There are two daughters growing up, preschool girls.

- And how old are they?

- Well-oo-oo … The product of their ages is equal to the number of pigeons under our feet.

- This information is not enough for me!

- The eldest is 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 old. Since the answer became clear only after receiving information that one of the daughters was older, it means that before that there was ambiguity. At first, based on the number of pigeons, the option was considered that the daughters are 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 is the age when children go to school, that is, they 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 to or less than 7, - 4 (1 × 4). Accordingly, the daughters are 1 and 4 years old. There are no other whole and at the same time "preschool" options.

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3. Where is my car ?

They say this task is given to junior high school students in Hong Kong schools. Children solve it literally in a matter of seconds.

What is the number of the parking space occupied by the car?

87. To guess, just look at the picture from the other side. Then the numbers that you now see upside down will take the correct position - 86, 87, 88, 89, 90, 91.

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4. Love in Kleptopia

Jan and Maria fell in love with each other, communicating only via the Internet. Ian wants to send Maria a wedding ring by mail - to propose. But here's the trouble: the beloved live in the country of Kleptopia, where any parcel sent by mail will certainly be stolen - unless it is locked in a box with a lock.

Jan and Maria have many locks, but they cannot send keys to each other - after all, the keys will also be stolen. How can Jan send the ring so that it will surely fall into Maria's hands?

Jan must send Maria the ring in a locked box. Without a key, of course. Maria, having received the parcel, must cut her own lock into it.

The box is then sent back to Jan. He opens his lock with his own key and again addresses the parcel with the only remaining locked lock to Maria. And the girl has a key to it.

By the way, this problem is not just a theoretical logic game. The idea used in it is the fundamental Seven Puzzles You Think You Must Not Have Heard Correctly in the cryptographic principle of Diffie - Hellman key exchange. This protocol allows two or more parties to obtain a shared secret using a communication channel unprotected from eavesdropping.

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5. Looking for a fake

The courier brought you 10 bags, each with a lot of coins. And everything is fine, but you suspect that the money in one of the bags is fake. All you know for sure is that real coins weigh 1 g each, and counterfeit ones weigh 1, 1 g. There are no other differences between the money.

Fortunately, you have an accurate digital scale that shows weights down to a tenth of a 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 calculate exactly in one weighing exactly which bag contains counterfeit coins and is there 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 pile of money weighs 55 g, then there are no fake ones in any of the bags. But if the weight is different, you will immediately understand what is the serial number of a bag full of fakes.

Consider: if the readings of the scales differ from the reference ones by 0, 1 - counterfeit coins in the first bag, by 0, 2 - in the second, by 0, 3 - in the third … by 1, 0 - in the tenth.

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6. Equality of tails

In a dark, dark room (you can't see it at all, and you can't turn on the light), there is a table on which 50 coins are lying. You cannot see them, but you can touch them, turn them over. And most importantly, you know for sure: 40 coins initially lie heads up, and 10 - tails.

Your task is to divide the money into two groups (not necessarily equal), each of which will have the same number of coins, tails up.

Divide the coins into two groups: one 40, the other 10. Now turn over all the money from the second group. Voila, you can turn on the light: the task is completed. If you don't believe it, check it out.

Let us explain the algorithm for literary mathematicians. After blindly dividing into two groups, this is what happened: the first had x tails; and in the second, respectively, - (10 - x) lattices (after all, in total, according to the conditions of the problem, lattices are 10). And the eagles, thus, - 10 - (10 - x) = x. That is, the number of heads in the second group is equal to the number of tails in the first.

We take the simplest step - turn over all the coins in the second pile. Thus, all coins-heads (x pieces) become coins-tails, and their number turns out to be the same as the number of tails in the first group.

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7. How not to get married

Once the owner of a small shop in Italy owed a large sum to a moneylender. He had no opportunity to repay the debt. But there was a beautiful daughter who had long been liked by the creditor.

- Let's do this, - the moneylender suggested to the shopkeeper. - You marry your daughter for me, and I forget about the duty as a relative. Well, hands down?

But the girl did not want to marry an old and ugly man. Therefore, the shopkeeper refused. However, the potential son-in-law caught the hesitation in his voice and made a new proposal.

“I don’t want to force anyone,” the moneylender said softly. - Let chance decide everything for us. Look: I will put two stones in the bag - black and white. And let the daughter pull out one of them without looking. If it is black, we will marry her and I will forgive you the debt. If white - I will forgive the debt just like that, without demanding the hand of your daughter.

The deal looked fair, and this time the father agreed. The usurer bent down to the pebble path, quickly picked up the stones and put them in a bag. But the daughter noticed a terrible thing: both stones were black! Whichever one she pulled out, she would have to get married. Of course, it was possible to catch the usurer of deception by taking out both stones at once. But he could have gone into a rage and canceled the deal, demanding the debt in full.

After thinking for a couple of seconds, the girl confidently stretched out her hand to the bag. And she did something that saved her father from debt, and herself from the need for marriage. Even the moneylender admitted the fairness of her act. What exactly did she do?

The girl pulled out a stone and, without having time to show it to anyone, as if accidentally dropped it on the path. The pebble immediately mixed with the rest of the pebble.

- Oh, I'm so clumsy! - the shopkeeper's daughter threw up her hands. - But that's okay. We can look into the bag. If there is a white stone left, then I pulled out a black one. And vice versa.

Of course, when everyone looked into the bag, a black stone was found there. Even the moneylender was forced to agree: this means that the girl pulled out the white one. And if so, there will be no wedding and the debt will have to be forgiven.

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8. Your code is confused …

You locked your suitcase with a three-digit code lock and accidentally forgot the numbers. But memory offers you the following clues:

682 - in this code one of the digits is correct and stands in its place;
614 - one of the numbers is correct, but out of place;
206 - two numbers are correct, but both are out of place;
738 - generally nonsense, not a single hit;
870 - one digit is correct, but out of place.

This information is enough to find the correct code. What is he?

042.

Following the fourth hint, cross out the numbers 7, 3 and 8 from all combinations - they are definitely not in the desired code. From the first hint, we find out that either 6 or 2 takes its place. But if it is 6, then the condition of the second hint, where 6 stands at the beginning, is not met. This means that the last digit of the code is 2. And 6 is absent in the cipher at all.

From the third hint, we conclude that the correct numbers of the code are 2 and 0. In this case, 2 is in the last place. So, 0 is on the first. Thus, the first and third digits of the code become known to us: 0 … 2.

Checking the second tip. Number 6 had been shallowed earlier. The unit does not fit: it is known that it is not in its place, but all possible places for it - the first and the last - have already been taken. Thus, only the number 4 is correct. We move it to the middle of the received code - 042.

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9. How to share a cake

And finally, a little sweet. You have a birthday cake, which must be divided by the number of guests - into 8 pieces. The only problem is that it needs to be done with just three cuts. Can you handle it?

Make two cuts crosswise - as if you want to divide the cake into four equal parts. And make the third cut not vertically, but horizontally, dividing the treat along.