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Martin Gardner - Mathematical puzzles and entertainment. Martin Gardner - Mathematical Puzzles and Entertainment Triangle from Orlov

Puzzles allow you to train the brain and always be prepared for non-standard situations.
To resolve a dilemma, sometimes you need to break your head, approach it from several different sides, learn all the options. But when everything converges, you get the true pleasure, delighted with your own fiction, resourcefulness.

We have collected a few puzzles with coins that can be solved by a cup of coffee with friends or simply distracting from business and take out the intelligence.

1. Triangle

Move 3 coins in such a way as to flip the triangle top down.

2. Five coins

Place 5 coins in two rows of 3 coins in each.

3. Eight coins

Place one coin so that you get two rows of 5 coins in each.

4. Triangle from Orlov

Overcoming each time three coins lying in one row, make all the coins lay an eagle up.

5. 10x5

Place 10 coins in five rows of 4 coins in each. Place coins one to another is prohibited.

6. Six coins in 2 rows

Figure 5, coins are located so that you can only find two segments with 3 coins on each. The task is to add only one coin and get four sections of three coins in each. Place coins one to another is prohibited.

7. Archaeologist

The archaeologist found an ancient coin labeled 10 year BC. He immediately realized that this was a fake. Why?

Answers:

1. Triangle

2. Five coins

3. Eight coins

4. Triangle from Orlov

1. Coupled bolts. Two identical bolts are lined with cutting (Fig. 109). Taking them to stronger the heads so that they could not turn around, circle several times one bolt around the other in the direction indicated by the arrows (turning in front of these thumbs, you can clearly imagine the motion of the bolts).

Will bolt heads:

  • close
  • diverge
  • stay at the constant distance from each other?

Using when solving the task, real bolts are not allowed.

2. Around the world flight. A group of aircraft is based on a small island. The tanks of each aircraft can accommodate as much fuel that it is enough for flights of half globe. When refilling in the air from the tanks of one aircraft in the other tanks, you can pump any amount of fuel. On the ground, refueling can only be performed on the island. For convenience, the problem is assumed that refueling on Earth and in the air occurs instantly, without loss of time.

What is the minimum number of aircraft that can provide a flight of one aircraft in a large circle, if we assume that the speed and fuel consumption of all aircraft are the same and all the planes are safely returned to their base?

3. Circle on a chessboard. The side of the cell on the chessboard is equal 4 cm. What is the radius of the highest circle, which can be carried out on a chessboard so that it takes place only on black cells?

4. Universal plug. In many collections, puzzles are explained how to cut a plug, which can be tightly plugged, round and triangular holes (Fig. 110). No less interesting to calculate the volume of such a plug. Suppose that the radius of its round base is equal to a unit of length, height - two units and that the edge in its upper part (two units in length) is strictly over one of the diameters of the base and in parallel to it. All parallel cork sections, the plane of which is perpendicular to the upper edge, have the type of triangles.

The plug surface can be considered as formed by direct connecting points of the upper, straight and lower edge having a circle shape. Each straight parallel to one of the planes perpendicular to the upper edge.

Of course, the volume of traffic jam is easy to calculate the methods of analysis, but it can be found and more simple way, knowing only that the volume of the straight cylinder is equal to the product of its base of its base.

5. Repeated number. If you gather guests, you can surprise them with an unusual focus. Ask one of the guests - I will call it a - write on a piece of paper some three-digit number twice in a row to get a six-digit number (for example, 394 394). Drink so that you cannot see the written number, and ask and transfer the piece of the guest to another guest in whom you ask to divide the number at 7.

"Do not worry about the rest, it will not be", "you say a guest in, and he is surprised to make sure that you are right (for example, 394 394 in division by 7 gives 56 342). Without telling you the result, in transfers the paper sheet to the third guest with, which divides the resulting in the result by 11. You again argue that the residue will not be, and again it is right (56 342 for division by 11 gives 5122).

Without turning around for guests and not knowing which numbers are written on a piece of paper, you ask to transfer it to the fourth guest d, which should divide the last result by 13. Again the division occurs without a balance (5122 during division by 13 gives 394). The final result D writes on a piece of paper and, folding it, gives you. Without unfolding a leaf with the answer, you pass it a and say: "Expand the sheet and you will see your three-digit number."

Prove that the focus is always obtained, no matter what number will select the first guest.

6. Clash rockets. Two rockets fly towards each other, one - at speeds 9000 miles / hourand the other - at speeds 21000 miles / hour. Their starting platforms are at a distance 1317 Mil One of the other: without taking this pencil and paper, count what distance will be between rockets per minute before collision.

7. How to move coins. On a flat smooth surface (for example, on the table), a triangle of six coins is posted (Fig. 111). It is required for the smallest number of moves to move the coins so that they formed the ring shown in Fig. 111 below. Each move consists in moving only one coin. It is impossible to shift from the place of other coins. In a new position, each coin must touch two other coins. It is not allowed to raise coins from the surface when solving the problem.

8. Handshakes and graphs. Prove that the number of participants in the last Congress of biophysicists, who exchanged handshakes an odd number of times, even. The same task allows and graphical interpretation. On the sheet of paper, put any number of points (each point depicts a member of the Congress). Between any two points is allowed to carry out any lines. Each point can unlimited number "exchange hands" with other points or be unavigible and not to greet anyone. Prove that the number of points from which an odd number of lines is coming is even.

9. Unusual duel. Smith, Brown and Jones, deciding to make some diversity into the usual duel on the guns, agreed to spend a match in several modified rules. I pull out the lot and learning, to whom it fell to shoot the first to whom - the second and to someone - the third, they diverged in their places, put in the tops of the equilateral triangle. We agreed that each in turn produces only one shot and can aim in anyone. Duel continues until any two participants are killed. The sequity of shooting is determined only by the results of the draw and remains unchanged during the entire duel.

All three participants know their opponents. Smith never woven, Brown falls into. Objective in 80% of cases, and Jones shooting worse than everyone, lashes as often as he falls into the goal.

Which of the duelists has a higher chance to survive, if we assume that all three adhere to optimal strategies And none of them will be killed by a crazy bullet, intended for another? A more difficult question: what is the probability of staying alive for each of the duelists?

Answers

1. The heads of the bolts do not come closer and do not diverge. Bolt movement can be compared with the movement of a person going up the descending escalator with the speed of the escalator.

2. To ensure round-the-world flight of one aircraft there are enough two aircraft. You can do this in many ways. The method offered by us seems to be the most economical: only five fuel stations are consumed, the pilots of the two aircraft providing flights have time to drink coffee and intercept the sandwich on the sandwich, and the entire method has not devoid of symmetry.

Airplanes A, B and C start at the same time. Flying 1/8 intended distance (i.e. the length of the circumference of a large circle), with patching 1/4 of the original stock of fuel in the tanks A and 1/4 - in the tanks in, after which it remains 1/4 of the refueling. This amount of fuel is enough to return to the base.

Airplanes A and B, continuing the flight, take another 1/8 of the round-the-world route, after which it pumps 1/4 refueling in Baki A. Baki in remain filled with exactly half, and it safely reaches the native airfield, while putting the landing with empty tanks.

Fully refilled plane A continues to fly until he ends up fuel. By this time, he is at a distance of 1/4 of the total path from the base, and he is met by a plane with who managed to repaid on the island. C pumped in tanks and 1/4 refueling and following and takes a course on the base. At a distance of 1/8 of the circumference of the globe, fuel and ends, but here they are encountered by visiting the base, which gives each of them 1/4 full refueling. After that fuel in the pots of each aircraft, it is enough for a safeguard to return to its base (though, it comes from empty tanks).

Graphically, the entire flight can be depicted using a diagram shown in Fig. 112, where the distance is delayed along the horizontal axis, and vertical - time. The right and left edges of the diagram should be considered glued.

Taking a circular solution to equal square root from 20 cm And putting it the edge into the center of the black cage on a chessboard with four-stage cells, you can describe the largest of the circles passing only on black cells.

4. Any cross-section of the plug by a plane, perpendicular to the upper edge and base, has a triangle type. If the cork was cylindrical, the corresponding sections would be rectangles, while the area of \u200b\u200beach rectangular section would twice the area of \u200b\u200bthe triangular cross section. Since the cylinder can be considered composed of rectangular cross sections, the volume of the universal cork should be half the volume of the cylinder: the volume of the cylinder is 2π, therefore, the volume of the universal plug is equal to π.

In fact, there are infinitely many traffic jams of various shapes that all three holes can be plugged. The plug of that form, which is described in the condition of the task, has the smallest volume compared to any convex body, capable of plusing the same three holes. The plug of the greatest volume is easy to obtain if it is shown to cut a cylindrical cork in the way. Fig. 113. This is this cork form usually mean compilers of puzzle collection, offering readers to find a universal plug suitable for round, triangular and square holes. Its volume is 2π - 4/3.

5. To write in a row two times the three-digit number - it's like multiplying this number per 1001. The number 1001 decomposes into simple multipliers 7, 11 and 13, so by ascribing it to the three-digit number of it once again to the right, which just integrates its number at 7 × 11 × 13. By dividing the six-digit number at 7, 11 and 13, it, naturally, receives a starting three-digit number again. This task is borrowed from the book Ya. I. Perelman * .

* (Ya. I. Perelman, Live mathematics, ed. 9, m, publishing house "Science", 1970.)

6. Two rockets come closer at speed 30 000 miles / hour, or 500 mph. Counting the time ago, from the moment of collision, we get that a minute before the rocket collision would have to be in a distance 500 miles Friend from each other.

7. Consider the initial location of the coins in the form of a triangle. Denote the number 1 upper coin, numbers 2 and 3 - coins in the next row and numbers 4, 5, 6 - coins in the bottom row. The following four strokes allow you to obtain an idea of \u200b\u200ba set of other solutions: Move the coin 1 so that it touches 2 and 4; coin 4 let's move so that it touches 5 and 6; The coin 5 will move so that it touches the coins 1 and 2 below, and finally coin 1 move so that it touches the coins 4 and 5.

8. Since two people participate in each handshake, the total number of handshakes, which exchanged all the participants of the Congress is shared by 2 and therefore even. The number of handshakes per share of those who exchanged with their colleagues an even number of handshakes are obviously even. Only the amount of an even number of odd components can be even number, therefore the number of those participants of the Congress, which exchanged with other participants an odd number of handshake, even.

The same statement can be proved otherwise. Before the start of the Congress, the number of its participants who exchanged an odd number of handshake is 0. After the first handshake, two "odd participants" appear. All handshakes, starting from the second, are divided into three types: Handshakes between two "even" participants, handshakes between two "odd" participants and "mixed" handshakes between "even" and "odd" participants. Each "even-even" handshake increases the number of "odd" participants on 2. Each "odd-odd" handshake reduces the number of "odd" participants also on 2. Each "odd-even" handshake turns the "odd" participant in the "even" and On the contrary, the "even" participant in the "odd" and, thus, leaves the number of "odd" participants unchanged. Therefore, the even number of biophysicists, who exchanged an odd number of handshakes, cannot change its parity and should always remain even.

Both evidence is applicable to the Count, on which the lines associate Points pairwise. Count lines form a network. The number of network points from which an odd number of lines is income, even. This theorem will meet us again in chapter 22 when considering puzzles associated with the wandering on the line of lines.

9. The greatest probability survive in the "triangular" duel is the worst of the shooters, Jones. Following him goes Smith, which never lashes. Since Jones opponents, when they come to shoot their turn, kiss each other, the optimal strategy for Jones is to shoot in the air until one of his opponents is killed. After that, he shoots the remaining enemy, having a big advantage in front of him.

It is easiest to calculate the probability of staying alive for Smith. In Duele with Brown with a probability of 1/2, he shoots first. In this case, he kills Brown. Brown, which falls into the target in 4 cases out of 5, shoots the first to also with a probability of 1/2. In this case, Smith remains alive with a probability of 1/5. Thus, Smith with a probability of 1/2 + 1/2 × 1/5 \u003d 3/5 is experiencing Brown. If Smith remains alive, then Jones shoots it, which in 1/2 of all cases lashes. But if Jones lashes at his first shot, then Smith, waiting for his turn to shoot, kills him. Therefore, with a probability of 1/2, Smith comes out of a duel with Jones alive and unharmed. So, the probability of staying alive after a duel with both of their opponents for Smith is 3/5 × 1/2 \u003d 3/10.

The case with Brown is more complicated, because it requires consideration of an infinite set of cases. The probability of staying alive after a duel with Smith for Brown is 2/5 (we just showed that Smith in Duele with Brown has the probability of a survival, equal to 3/5; since only one of the duelists should remain alive, the probability for Smith We find, deducted 3/5 out of 1). Then he shoots Jones in Brown, which falls into the goal only in half cases. If Jones wipes, then Brown with a probability of 4/5 kills him. So, at this stage, Duele Brown with a probability of 1/2 × 4/5 \u003d 4/10 comes out the winner of the fight with Jones. But with a probability of 1/5, Brown may miss, after which Jones has the right to shoot again. With a probability of 1/2, Brown will remain alive, and then he in turn will be able to shoot in Jones and with a probability of 4/5 kill him. Braun's chances of staying alive during the second round of the duel are 1/2 × 1/5 × 1/2 × 4/5 \u003d 4/100.

If Brown misses again, then during the third round he can kill Jones only with a 4/1000 probability. In the case of re-miss during the fourth round, it will fall into Jones with a probability of 4 / 10,000, etc. Thus, Brown's chances to survive Jones are equal to the sum of the infinite row

This is nothing more than an infinite periodic decimal fraction 0,44444 ..., equal to 4/9.

Previously, we saw that Brown with a probability of 2/5 can survive Smith. Just we have shown that with a probability of 4/9, he will survive after a duel with Jones. The likelihood that it is Brown who will survive both of their opponents is, therefore, 2/5 × 4/9 \u003d 8/4.

The whole duel is conveniently portrayed using a special graph - Duel tree (Fig. 114). At first, the trunk of the tree is split. This is because if Jones shoot the first, he produces his shot into the air, after which two equivalent opportunities remain: either Smith shoots, or Jones (these two shoot "quite seriously", with the solid intention to kill their opponent). One of the branches of the tree extends to infinity. Calculation of the probability for a particular duelist remains alive as follows:

  1. It is necessary to note all the branches of the tree in which the participant of the party you are interested in is the only one of all the three remaining.
  2. Walking from each of the marked branches back to the root of the tree, you should multiply the probabilities of all passed segments of the path. The work will give the likelihood of an event corresponding to the end of the marked branch.
  3. Fold all the probabilities calculated in paragraph 2. Their amount will be interested in us by the probability of survival of a particular duelist.

When calculating the probability, survive for Brown and Jones have to take into account the infinitely many branches, however, with the help of a graph, it is not difficult to specify the formula of the general member of the relevant series.

Various options for this task are included in many collections of puzzles.

Put it in front of ourselves 3 two-fiber and 2 tenth-inextric coins as shown on the top figure. Now try moving them so that they take the position shown below. Of course, it is necessary to spend as few times as possible. Under the condition, every time you need to move at the same time 2 next to the underlying coins: one of them should be definitely two-achievable, and the other tenthly agreary.

During the movement of the coin, it is impossible to separate one of the other, it is impossible to change them and places. In other words, that coin, which at the beginning of the move was on the left, should remain on the left. During the permutations, the chain of coins may arise, but after the last permutation, the chain must become solid again. After the last changes, the coin does not have to be on the same place they occupied at the beginning. The task is not as simple as it may seem at first glance.

The task is solved in 4 stroke:

  1. Move the third and fourth coins (the numbering is given to the left right) to the right so that between them and the fifth coin was a rupture of 2 coins width.
  2. Move the first and second coins to the right for the displaced third and fourth coins so that the first and fourth coins concern one other.
  3. Move the first and fourth coins into the gap between the fifth and the third.
  4. Move the fifth and fourth coin into the interval between the third and the second.


"Your Free Time", V.N. Bolovitinov, B.I. Kolovtov

Near a small white house on the edge of the forest two weeks ago at 23.30 there was a corpse of some L. Clematner. The investigation went into a dead end, and, as always, in such cases, the inspector of the breast was called to the aid. - This happened as follows, - began to tell witness M. during inspection by the inspector of the scene. - I rode a bike along the edge ...

With the onset of the protracted autumn rain, the Vimmer leaned and returned to his city apartment. Even before moving, the Vimmer agreed with a neighbor in the country of some kind, so that he looked after his economy. A few days after the new year, Z. called the vimmer and the excited voice reported that the cottage was robbed. The vimmer immediately appealed to the police, and ...

- Here, everything happened on the edge, Mr. Inspector. Some guy threw me in the face of ground peppers and snatched a portfolio from the hands, where there was money in the amount of a thousand eight hundred fifty-seven marks. I shouted, but there was no one around. From terrible burning in my eyes I have not seen anything. Fortunately, I know this area ...

In one of the villas on the edge of the city 3. Some Alfredo di Meiro settled. Thanks to pleasant outfit and aristocratic mannera, he managed to achieve the location of the most influential people of the city. Using extraordinary popularity in the city, Don Alfredo held money to the right and left. When the amount of his debts accepted impressive sizes, Villa Alfredo di Meiro visited the inspector of the Vernik. Master…

- I have to carefully search your apartment, as your neighbor claims that yesterday, on New Year's Eve, while he, together with your family, sowed funny songs under your Christmas tree, sparkling multicolored light bulbs, you penetrated his apartment and kidnapped him A number of valuable things, - said the inspector of the Vernik, turning to Mr. Mayer. - ...

- Yes, you will finally leave me! I am very hurry. - If you do not stop right now, I will have to resort to extreme measures, "said the inspector of the breastnik, as ever, by the way, turned out to be at the scene. - After all, you have now taken gloves in the store and left, without paying for them! - It is not true! I really only ...

During the service on the affairs of the service in California, the inspector of the vannik was introduced to demonstrate their uncommon abilities to local police. Once he was urgently called at the airfield, located near Los Angeles. From the accidentally heard telephone conversation of the police, it became known that the flight of the aircraft, which is sent to Alaska, will end with a disaster. One of the passengers, intending to commit suicide, takes with him ...

Yesterday I returned home from the service somewhat earlier than usual. Only I sat down at the table, going to dine, how suddenly something fell into the wife's room. I rushed there and saw the old vase lying on the floor, which my wife is very valuable. Vase was broken. At the same time, some kind of man ran out of the room. I rushed after him ...

A dangerous criminal ran away from prison. For a long time he managed to hide, but in the end, the inspector of the vannik attacked his trail. In one village, he and his companion said that a truly unknown was held here 15 minutes ago and headed towards the field. Yes, he's as if she looks like a person depicted in the photo, which he kept ...

In the hotel "Grune Tane" everything was ready for great reception. The waiter last went down to the basement for the last time to check whether wine was enough. And suddenly ... oh horror! Half of wine reserves disappeared. Inspector Vernik, who turned out to be completely accidentally in this city, immediately arrived at the hotel and, not bag, began to investigate the case. It…

4 coins

On the table lie 4 coins. They must be shifted from the first position to the second, while the place in the middle must accurately correspond to one coin. By moving coins, you can not tear them from the table, and you can move one coin only to two or three others. In this and the following tasks, it is forbidden to use measuring instruments - ruler, circulation, etc. All measurements are only with coins.

ANSWER

Flower

Flower is complicated out of 7 coins. It is necessary for the minimum number of movements from the center of the flower to get a coin without changing its shape and without taking coins from the table.


ANSWER

Triangle

Move 3 coins in such a way as to turn the triangle top up.


ANSWER

Parallelogram

For the minimum number of displacements, turn the parallelogram into the triangle.


ANSWER

Hexagon

For the minimum number of displacements, turn the hexagon into the triangle.


ANSWER

Two unnecessary coins

For the minimum number of movements, we will remove the coins from the center 2 without changing the shape of the shape and without taking the coins from the table.


ANSWER

Three coins in a triangle

For the minimum number of movements, get from the center of 3 coins, without changing the shape of the shape and without taking the coins from the table.


ANSWER

Five coins

Place 5 coins in two rows of 3 coins in each.


ANSWER

Six coins

Place 6 coins in two rows of 4 coins in each.


ANSWER

Six coins-2

Figure 5, coins are located so that you can only find two segments with 3 coins on each. The task is to add only one coin and get four sections of three coins in each. Place coins one to another is prohibited.

ANSWER

Eight coins

Place one coin so that you get two rows of 5 coins in each.


ANSWER

Twelve coins

Place 12 coins in seven rows of 4 coins in each.


ANSWER

Ten coins

Place 10 coins in four rows of 4 coins in each.


ANSWER

4x4

On the table lie sixteen coins - four rows of four coins. Remove the six coins so that in each line and in each column there is an even number of coins.


ANSWER

10x5

Place 10 coins in five rows of 4 coins in each. Place coins one to another is prohibited.


ANSWER

12x6

Place 12 coins in six rows of 4 coins in each. Place coins one to another is prohibited.


ANSWER

Rhombus

There are 2 coins of 1 ruble, 2 for 2 rubles and 2 to 10 rubles. Place all six coins in the form of rhombus so that coins of one dignity do not come into contact with each other.


ANSWER

Perimeter of triangle

There are 5 coins on 1 ruble and 5 to 2 rubles. Place all ten coins in the form of a triangle so that the amount of numbers on each of its three sides is the same.


ANSWER

Moving 8 coins

There are 8 coins (4 coins of 1 ruble and 4 coins of 2 rubles). All coins are placed in one row. By moving in one run two next to the underlying coins, for 4 strokes, so that one-bored and double-click coins alternate.


ANSWER

Moving 6 coins

There are 6 coins (2 coins of 1 ruble, 2 for 2 rubles and 2 to 10 rubles). All coins are placed in one row. By moving in one run two next to the lying coins, for 3 strokes, do the way shown in the figure below.


ANSWER

Square coin

In the square 3x3 around the perimeter lies 8 coins (4 coins of 1 ruble and 4 coins of 2 rubles). Using a free cage in the center, change places one-boiled coins and double-bubbles. One step is allowed to move one coin into a free adjacent horizontal cell or vertical. It is forbidden to put in one cell more than one coin, jump over the cells and go beyond the square.


ANSWER

Corners

Change places one-bought coins and double-clicking, using a free cage in the center. One step is allowed to move one coin into a free adjacent horizontal cell or vertical. You can jump over one cell with a horizontal or vertical coin. It is forbidden to put more than one coin in one cell, and go beyond the figure.


ANSWER

Five Orlov

Overcoming each time three coins lying next to, get 5 coins lying up.


ANSWER

Seven Orlov

Outping each time three coins lying nearby, get 7 coins lying up.


ANSWER

Triangle from Orlov

Overcoming each time three coins lying in one row, make all the coins lay an eagle up.


ANSWER

Eagles in a circle

Eight coins lie in a circle. Overting every time two coins lying in one row, make it so that all coins lay an eagle up.


ANSWER

Square from Orlov

Eight coins lie around the perimeter of the square. Overcoming each time three coins lying on any of his sides, make it so that all coins lay an eagle up.


ANSWER

Coins on the table (puzzle on the mixture)

On the table there are two coins - 1 ruble and 2 rubles. How to put 1 ruble coin under 2 ruble, if 2 ruble coins can absolutely touch and move off.

1 ruble Put on the floor under the place,

where on the table lies 2 rubles.

Hole

Cut on a sheet of paper a round hole slightly smaller than a double coin. Now, try to push a five-log coin into it! It is impossible to tear or handle paper, make manipulations with a coin (bending, told, etc.) is also prohibited.


ANSWER

Circular motion

Put the coin on the table in dignity of 5 rubles. After that, next to the second coin of 5 rubles and ride it (without slip) around the first coin. If the second five rubles rolled one full revolution, how many revolutions around their axis did he do?


Answer (Highlight text below)

Most likely, you will get one turn - you can not avoid

sliphes. But if it were not - there would be two turns!

Coins on a glass

Take 2. ordinary coins And one glass. Now install these coins on the edges of the glass. And now try to remove from the glass to the table with two fingers of one hand, and it is impossible to touch the glass. How to do it.


Answer (Highlight text below)

Put the big and index fingers of one hand on both coins and, pressing the outer walls of the glass, gently slide down - until you reach a glass of glass. And now the rapid movement take off the coin from the glass.

Jumping

Place eight coins in a row on the table, as shown in the figure, and then try in four strokes (coin moving) rebuild a row into four stacks of two coins. Another condition: at each move, the coin should "jump out" exactly through two coins (in any direction) and land on the nearest single coin. At the same time, it does not matter whether coins are lying through which they jump, close or each other.


Three in the row

On the table lies twelve coins. Coins lie in four rows of three coins. Coins alternate, part lies up an eagle, and a part of a wide. Your task, touching only one coin, move it in such a way that in all horizontal rows of the coins lay either an eagle or a wide-up. At the same time, coins should remain in the form of the same figure. Dump coins from the table are also prohibited.

Answer (Highlight text below)

Hand with a finger we move the middle coin of the top row and left. Without taking off the finger from the coin moving it down, ribbing the left column of coins until it turns out to be under the middle of the bottom row. Pushing this coin the middle column upwards until you get four rows of three identical coins in each. The task is solved, and you all touched only one coin.

Archaeologist

The archaeologist found an ancient coin labeled 10 year BC. He immediately realized that this was a fake. Why?


Answer (Highlight text below)

Designation "BC" Could appear only after the onset of our era.

Checkers with coins

There is a field of 5x5 cells. Each coin can jump over one other horizontally or vertical. A coin through which another coin was thrown into - removed from the field. How to remove all coins except one that should stay in the center of the field?


ANSWER

9 coins

The figure below shows how 16 coins of 4 coins can be positioned in each of the 10 rows (4 vertical, 4 horizontal and 2 diagonal). And how can I arrange 9 coins in 10 rows of 3 coins in each row?


ANSWER

Square from the triangle

The triangle is posted in the pattern of coins. How to move only two coins to get a square?


ANSWER

Overvents

9 coins lie on the square in the form of a square. Some of them lie up, some of them. In one move, the coin turns over the coin with the other side, as well as all coins in contact with it horizontally and vertical. Below are examples of moves. Try to turn over all coins with eagles over 3 stroke.


1. Coupled bolts. Two identical bolts are lined with cutting (Fig. 109).

Fig. 109.Coupled bolts.

Taking them to stronger the heads so that they could not turn around, circle several times one bolt around the other in the direction indicated by the arrows (turning in front of these thumbs, you can clearly imagine the motion of the bolts).

Will bolt heads: a) get closer, b) to disperse or c) stay at a constant distance from each other?

Using when solving the task, real bolts are not allowed.

2. Around the world flight. A group of aircraft is based on a small island. The tanks of each aircraft can accommodate so much fuel that it is enough for the flight half of the globe. When refilling in the air from the tanks of one aircraft in the other tanks, you can pump any amount of fuel. On the ground, refueling can only be performed on the island. For the convenience of solving the problem, it is assumed that refueling on Earth and in the air occurs instantly, a time loss of time.

What is the minimum number of aircraft that can provide a flight of one aircraft in a large circle, if we assume that the speed and fuel consumption of all aircraft are the same and all the planes are safely returned to their base?

3. Circle on a chessboard. The side of the cell on a chessboard 4 cm. What is the radius of the highest circumference, which can be carried out on a chessboard so that it passes only in black cells?

4. Universal plug. In many collections, puzzles are explained how to cut a plug, which can be tightly plugged, round and triangular holes (Fig. 110).

Fig. 110.Universal cork.

No less interesting to calculate the volume of such a plug. Suppose that the radius of its round base is equal to a unit of length, and the height is two units and that the edge in its upper part (two units in length) is strictly over one of the base diameters and in parallel to it. All parallel cork sections, the plane of which is perpendicular to the upper edge, have the type of triangles.

The plug surface can be considered as formed by straight, connecting points of the upper, straight and lower, having a circle shape, ribs. Each straight parallel to one of the planes perpendicular to the upper edge.

Of course, the volume of the traffic jam is not difficult to calculate the analysis methods, but it can be found in a simpler way, knowing only that the volume of the direct cylinder is equal to the product of its base to height.

5. Repeated number. If you gather guests, you can surprise them with an unusual focus. Ask one of the guests - let's call it a - write on a piece of paper some three-digit number two times in a row to get a six-digit number (for example 394 394). Drink so that you can not see the written number, and ask and transfer the sheet to another guest, in whom you ask to divide the number 7.

"Do not worry about the rest, it will not be", "you say a guest in, and he is surprised to make sure that you are right (for example, 394 394 in division by 7 gives 56 342). Without telling you the result, in transfers the paper sheet to the third guest, with, which divides the resulting on the result by 11. You again claim that the residue will not be, and again it is right (56 342 for division by 11 gives 5122).

Without turning around for guests and not knowing what numbers are written on a piece of paper, you ask to transfer it to the fourth guest, D, which should divide the last result by 13. Again the division occurs without a balance (5122 during division by 13 gives 394). The final result D writes on a piece of paper and, folding it, gives you. Without unfolding a leaf with the answer, you pass it a and say: "Expand the sheet and you will see your three-digit number."

Prove that the focus is always obtained, no matter what number will select the first guest.

6. Clash rockets. Two rockets fly towards each other, one - at a speed of 9000 miles / hour, and the other - at a speed of 21,000 miles / hour. Their start-up sites are at a distance of 1317 miles one from the other. Do not use the pencil and paper, calculate what distance will be between rockets per minute before collision.

7. How to move coins. On a flat smooth surface (for example on the table), a triangle of six coins is laid out (Fig. 111).

Fig. 111.How to move coins?

It is required for the smallest number of moves to move the coins so that they formed the ring shown in Fig. 111. Each move is to move only one coin. It is impossible to shift from the place of other coins. In a new position, each coin must touch two other coins. It is not allowed to raise coins from the surface when solving the problem.

8. Handshakes and graphs. Prove that the number of participants in the last Congress of biophysicists, who exchanged handshakes an odd number of times, even. The same task allows and graphical interpretation. Place any number of points on the paper sheet (each point depicts a member of the Congress). Between any two points is allowed to carry out any lines. Each point can unlimited number "exchange hands" with other points or be unavigible and not to greet anyone. Prove that the number of points from which an odd number of lines is coming is even.

9. Unusual duel. Smith, Brown and Jones, deciding to make some diversity into the usual duel on the guns, agreed to spend a match in several modified rules. I pull out the lot and learning, to whom it fell to shoot the first to whom - the second and to someone - the third, they diverged in their places, put in the tops of the equilateral triangle. We agreed that each in turn produces only one shot and can aim in anyone. Duel continues until any two participants are killed. The sequity of shooting is determined only by the results of the draw and remains unchanged during the entire duel.

All three participants know that Smith never lashes, Brown falls into a goal in 80% of cases, and Jones shooting worse than everyone, wakes up as often as it falls into the goal.

Which of the duelists has a higher chance to survive, if we assume that all three adheres to optimal strategies and none of them will be killed by a crazy bullet intended for another?

A more difficult question: what is the probability of staying alive for each of the duelists?

Answers

1. The heads of the bolts do not come closer and do not diverge. Bolt movement can be compared with the movement of a person going up the descending escalator with the speed of the escalator.

2. To provide around the world of one aircraft, there are enough two aircraft. You can do this in many ways.

The method offered by us seems to be the most economical: only five fuel stations are consumed, the pilots of the two aircraft providing flights have time to drink coffee and intercept the sandwich on the sandwich, and the entire method has not devoid of symmetry.

Airplanes A, B and C start at the same time. Flying 1/8 of the outlined distance (i.e. the lengths of the circumference of a large circle), with the 1/4 of the original stock of fuel in the tanks A and 1/4 - in the tanks in, after which it remains 1/4 refueling. This amount of fuel is enough to return to the base.

Airplanes and B, continuing flight, pass another 1/8 round-the-world route, after which it pumps 1/4 refueling in Baki A.

Tanks in remain filled with exactly half, and it safely reaches the native airfield, putting the landing already with empty tanks.

Fully refilled plane A continues to fly until he ends up fuel. By this time, he is at a distance of 1/4 of the entire path from the base, and he is met by a plane with who managed to repaid on the island. C pumped in tanks and 1/4 refueling and following and takes a course on the base. At a distance of 1/8 of the circumference of the globe, fuel and ends, but here they are encountered on the basis of the Base in, which gives each of them to 1/4 full refueling. After that fuel in the pots of each aircraft, it is enough for a safeguard to return to its base (though, it comes from empty tanks).

Graphically, the entire flight can be depicted using a diagram shown in Fig. 112, where the distance is delayed along the horizontal axis, and vertical - time. The right and left edges of the diagram should be considered glued.