9 logical tasks that the teeth only intellectuals

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July 16. This follows from the dialogue that took place between Albert and Bernard. Plus a little process of elimination. See.

If Cheryl was born in May or June, so her birthday can be 19th or 18th. These numbers are found in only one time list. Accordingly, Bernard, hearing them, just could see about a month in question. But Albert, as follows from its first replica, I'm sure Bernard, knowing the number, definitely will not be able to call a month. So, we are not talking about May or June. Cheryl was born in the month, each of these dates in which there is a double in the adjacent months. That is - in July or August.

Bernard, who knows of birth, hearing and analyzing replica Albert (that is, finding out about July or August), says that now knows the correct answer. From this it follows that a certain number of Bernardo - not 14, because it is duplicated, and in July and August, so it is impossible to determine the correct date. But Bernard confident in his decision. So, the number known to him does not double in July and August. Under this condition get three options: 16 July, 15 August and 17 August.

In turn, Albert heard the words of Bernard (and logically before reaching the above-mentioned three possible dates), stated that now also knows the correct date. We remember that Albert is known for a month. If this month was August, the young man could not determine the number of - in fact, in August featured just two. So, there is only one option - 16 July.

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1 and 4 years. Since the answer became clear only after receiving information that one of the daughters of older, so before there was ambiguity. At first, based on the number of pigeons, the variant that the daughters - twins (that is, their ages are equal). This is only possible with the number of pigeons is the square numbers up to and including 7 (7 years old - the age when children go to school, that is, cease to be preschoolers): 1, 4, 9, 16, 25, 36, 49.

From these squares, only one can be obtained by multiplication of two different numbers, each of which is equal to or less than 7, 4 - (1 × 4). Accordingly, the daughters of 1 and 4 years. Other whole while "pre" There is no alternative.

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

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Yang should send Maria a ring in a locked box. Without the key, of course. Maria received the parcel, should be embedded in her own castle.

Then Jan sent a box again. He opened his own key and lock re-addresses the parcel with the only remaining locked lock Mary. And the girl is the key to it.

Incidentally, this task - not just a theoretical game of logic. Used in its idea - a fundamentalSeven Puzzles You Think You Must Not Have Heard Correctly Key Exchange cryptographic principle of Protocol Diffie - Hellman. This protocol enables two or more parties to get the shared secret key over an insecure communications channel listening.

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One weighing enough. Just put on the scales just 55 coins: 1 - from the first bag, 2 - in the second, 3 - in the third, four - from the fourth... 10 - from the tenth. If the whole bunch of money will weigh 55 g - so fake is not in one of the bags. But if the weight will be different, you will immediately understand what the serial number of bags full of fakes.

Consider if indications of weights will differ from the reference 0.1 - counterfeit coins in the first bag, 0.2 - in the second, 0.3 - in the third... 1.0 - a tenth.

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Divide the coins into two groups: one 40, the other 10. Now turn all the money from the second group. Voila can include light: the task is completed. Do not believe me - check.

For pedants mathematicians explain the algorithm. After the blind separation into two groups here is what happened: the first left's tails; and second, respectively, - (10 - x) tails (for a total of 10 conditions tails problem). A eagles thus - 10 - (10 - x) = x. That is, the number of eagles in the second group is the number of tails in the first.

Do simple step - turn over all the coins in the second bunch. Thus, all coin-eagles (pieces) are coins-tails, and their number is the same as the number of tails in the first group.

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She pulled out a stone, and without having to show anyone, if accidentally dropped it on the track. Stone then mixed with the rest of the pebbles.

- Oh, I'm so clumsy! - clasped hands shopkeeper's daughter. - But that's nothing. After all, we can look in the bag. If there was a white stone, so I pulled the black. And vice versa.

Of course, when all looked in the bag, there was discovered a black stone. Even moneylender I had to agree: it means that the girl pulled the white. And if so - the wedding will not be and have to forgive the debt.

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042.

Following the fourth tip, cross out all of the combinations of the numbers 7, 3 and 8 - they are in the desired code is definitely not. We find out from the first hint that his place is either 6 or 2. But if it is 6, it is not the condition of the second clue where 6 stands at the beginning. So, the last digit of the code - 2. A 6 is absent in the cipher.

From the third clue we conclude that the right numbers of the code - 2 and 0. In this case 2 is at the bottom. This means that the 0 - on the ground. Thus, we become aware of the first and third digits of the code: 0... 2.

Is compared to the second clue. Figure 6 shallows before. The unit does not fit: it is known that it is not on its place, but all the possible locations for it - first and last - are already occupied. Thus, the true figure is only 4. It and move into the middle of the resulting code - 042.

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Make two cuts crosswise - as if you want to divide the cake into four equal parts. A third incision swipe not vertically but horizontally along the dividing treat.