Kvant 1970, No. 5 · Pages

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IN THE ROOM:

What cars can do 2
Length measurement 10
Results of the final round
Weight optimization of schoolchildren 17
Geometry tabletop 18
Comparisons by model and residual arithmetic 27
Elementary theory of flight and water waves 34
Problem book “Quenta” 39
How to grow crystal 42
On entrance exams in mathematics in technical schools in 1969 45
Someone loses, someone finds 48
TV is your assistant when preparing for VZZ! 52
Boiling water and frost 54
Cylindrical chess 56
Textbook “Mathematical Analysis” 58
Answers, directions, solutions 60
Problems for 5th grade 64
Crossword “Geometric” -

3rd cover page

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running_head:
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R. S. GUTER
WHAT
UMENOT
CARS

Different times - different ma-
tires are different for different cars
possibilities. Apparently the first

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remembers the matrix of a modern computer.
It’s easy to verify this by comparing the illustration of Swift’s machine that we borrowed from the Russian edition of 1947 (and, presumably, reproducing the corresponding Swift original) with a photograph of the matrix of the memo, taken ten years later.

But if “logical machines” could be referred to as manufacturing, then the need for computers that would help people count was clear, and that’s why they were created quite a long time ago.

Apparently, counting began “on the fingers,” so fingers should be considered the first (or primitive) computing machine. The next step of the previous computing machines should be considered the calculating abacus and the Russian abacus, which has been brought to this day, where numbers were represented by pebbles or bones put on knitting needles. The calculations on these devices are done entirely by hand, so the abacus should be used as a device to facilitate calculation, but not as a calculating machine.

The history of computing machines begins in 1642, when Blaise Pascal invented the tenth counting wheel and a mechanism for automatically transferring from rank to rank. The idea of ​​this transfer mechanism is still used today in desktop keyboard machines, essentially without any changes. But the level of technology at that time was too low to create reliably working mechanisms and devices. It took about another hundred and fifty years to implement these ideas and create practical usable computers.

In 1874, the Russian engineer V. T. Odner invented and built an arrithometer, which used gears with a variable number of teeth to display numbers. In 1878, the great Russian mathematician P. L. Chebyshev (1821-1894) invented an arithometer with a continuous transfer mechanism from rank to rank. In 1891, industrial production of Odhner arithometers began. The success of this design is evidenced by the fact that the production of manual Felix arithometers, essentially no different from Odner’s arithometers, was only replaced in the mid-fifties by the production of electromechanical arithometers.

The adding machine has also become widespread.

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The main characteristics are the speed of performing arithmetic operations and the amount of memory. The first Soviet electronic computers BESM, M-2, Strela, Ural had a speed of 200-2000 arithmetic operations per second and a memory capacity of 1024-2048 cells. These machines were tube machines and had several thousand vacuum tubes each.

Nibaulier distributed machines of the late sixties are BESM-4, M-220, Minsk-22, Razdan-3 with speeds of about 25,000 arithmetic operations per second and a RAM capacity of 8192-32,768 cells. The BESM-6 machine, with a RAM of 32,768 cells, already has a speed of up to a million arithmetic operations per second. In these machines, vacuum tubes have already been completely replaced by semiconductor elements (transistors), which have a significantly smaller volume and require significantly less power.

The use of transistors has made it possible to create small-sized desktop electronic keyboard machines that replace electromechanical ones. They are distinguished by high speed and complete noiselessness, which is very important for large computing laboratories. Therefore, desktop keyboard electronic machines are beginning to replace electromechanical ones. Our cars “Vyatka”, “Iskra”, “Orbita” and some others have gained wide popularity.

The first successes in the creation of computers made it possible to use them for new special tasks of various kinds.

It is not very difficult, at least theoretically, to “build” a computer into a control one. Let’s consider the problem of controlling some technological process, for example, the process of producing a chemical substance in a special reactor.

The course of the process is determined by the state inside the reactor: the concentration of the starting substances and the resulting substance, temperature and pressure inside the reactor, etc. Process control usually consists of changing controlled parameters, such as feed rate and temperature. In manual control, the operator, based on instrument readings, determines the state of the process at a given moment and, having determined the values ​​of the adjustable parameters that are most advantageous for a given state, brings them to the required values.

To automatically control such a process, it is necessary to create a computer with changing devices characteristic

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terizing the state of the process, and control devices. Having received information about the state of the process from the changers, the machine will expand the most advantageous mode for a given state and move the control devices to the required positions, and such calculation and adjustment can be repeated after certain periods of time.

Of course, in the practical implementation of such a scheme, a variety of difficulties can and do occur, but overcoming them is already, as they say, “a technical matter.”

Particularly important is the use of high-speed electronic computers to control high-speed processes. Many of these processes require regulatory movement tens or hundreds of times per second; Without taking into account high-speed machines, regulation of such processes is generally impossible. But this topic deserves a more detailed discussion and we are going to do this in one of the subsequent issues of the magazine.

The various “non-computational” applications of modern computers are quite wide and varied. It is very interesting, for example, to use them to translate texts from one language to another. We can give here only an approximate description of the principles of operation of such a program. But even from such a rough approximation one can imagine sufficient results and existing difficulties.

Let’s imagine two languages ​​consisting of the same number of words that do not change with declension or conjugation and all are meanings (in Russian, almost all words are meanings; for example, in other European languages ​​there are a number of service words that do not have independent meaning. These include, for example, der or ein in the meaning of the article in German, the - in English, etc.). Let us further assume that every word in one language has exactly one translation into another language. In mathematical terms, this means that there is a one-to-one correspondence between the words of the two languages ​​in question.

If we also assume that the order of words in a sentence is unimportant for both languages ​​(this is roughly how the situation is in the Russian language, where words in a sentence can be rearranged quite arbitrarily, unlike many others, where their order is prescribed by grammatical construction and the meaning of the written phrase can change from rearranging the words), then for such languages ​​it is quite simple to write a translation program.

It is enough to carry a dictionary into the car. This can be done, for example, in this indirect way. Zanuterum from the word of the first language in some way, for example alphabetically. The words of the second language are numbered by the same numbers so that the corresponding words of the two languages ​​have the same number. Now all that remains is to create a program with which the machine could find its number based on the entered word of the first language, then by the number corresponding to the word in the second language, and print it out. This will create a translation program.

For actual existing languages ​​the situation is, of course, much more complicated. First of all, words are usually inflective and the order of words in a sentence is not indifferent. Therefore, in addition to the dictionary, the program must contain a fairly detailed list of grammatical rules for both languages, which would allow:

1. analyze the entered sentence syntactically, determine the subject, predicate and

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It may have fewer words and less variety to translate each word. In addition, the syntactic structures of sentences in this case are less varied, which also makes the work easier. We have already made several such “narrow” programs for translation from French or English into Russian.

Somewhat later than work on machine translation, research began in the field of machine document, i.e., the use of computers to document theorems or, more correctly, to deduce one of the other statements from others. At the same time, I decided not only about already known statements.

At the international symposium on heuristic programming in February 1967, the American mathematician G. Gelernter spoke about his compilation of a program for documenting theorems of elementary geometry. Working with this program, the document machine

la theorem, the validity of which was previously unknown to the programmer.

The proven theorem is formed as follows. From a vertex of a square that does not lie on its diagonal, two segments are drawn connecting the vertex with the midpoints of the opposite sides. It is stated that these segments divide the diagonal into three equal parts (see figure). Of course, this theorem is not too difficult, and, presumably, readers will be able to prove it on their own. But this proof will not be the first - for the first time this theorem was proved by an electronic computer.

In the USSR, some programs of this kind have also been created, in particular, programs for proving theorems of arithmetic, allowing (which should be especially interesting for ninth grade students) to use the method of mathematical induction.

Even more interesting programs with the help of which electronic computers can play card games, dominoes, checkers and chess. We will not talk here about the achievements available in this area, intending to devote a separate article to these issues. In the same way, the use of modern computers in recognition and diagnostic problems deserves a more detailed analysis. We’re going to do this a little later.

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exchange and for all nations." The system was based on two basic units: the meter and the kilogram, with derivatives and decimal divisions. The simplicity of the metric system and the ease of its use fully satisfied the requirements of that time. The system was adopted by many countries, including Russia.

One forty millionth of the earth’s meridine passing through Paris was adopted as a unit of length - the meter. At the end of the 18th - beginning of the 19th centuries, a group of scientists, on behalf of the French Academy of Sciences, worked out the length of the meridine segment from Dunkirk to the island of Formentura *). Based on the measurements taken, a

*) The very first calculation of the circumference of the globe belongs to Eratosthenes (276-194 BC). He knew that in Siena there is a well, which is illuminated by the Sun on the day of the summer solstice, that is, in Siena at this time the Sun is at its zenith. At the same time, in Alexandria, where Eratosthenes lived, the Sun stood south of the plumb line at 1/50 of the circle, that is, at 7°12′. Eratosthenes determined the distance between Syene and Alexandria at 5000 stadia. From these data, he found the circumference of the globe, which Eratosthenes found equal to 39,690 km (in modern units), which differs slightly from the actual value.

Fig. 1. So in the Middle Ages they established what a foot is equal to.

prototype of the meter, which is still kept at the International Bureau of Weights and Measures in Sevres near Paris.

The meter was conceived as a natural unit of length - one forty-millionth of the earth’s meridine. But over time, it ceased to be such a unit. The circumference of the Earth was clarified, the platinum rod was changed, and now the sample meter is not one forty millionth of the circumference of the globe, but simply some fixed length. Therefore, there was a need to return to a natural standard that occurs in nature and is easily reproducible. Otherwise there would be a terrible mess if suddenly the international meter stage was lost, cleaned or damaged for some reason. In addition, the existing stage did not provide length measurements with the accuracy required by science and technology in need.

XI General Conference on Meters and Weights (1960) for a new definition of the meter. Its resolution reads: “…the conference, taking into account that the international prototype does not define the meter with precision sufficient for modern needs, and that, on the other hand, it is desirable to adopt a natural and

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Fig. 2. Comparison of the length of the reference meter with the length of the light wave is carried out using a device called interference comparators. One of the diagrams of such a device is shown in figure. A parallel beam of light rays from a source S falls at an angle of 45° onto a silvered translucent glass plate P, which divides the light flux into two parts. Reflected ray 1 goes to the mirror 3₁, is reflected from it and, having cleared the plate P₁, falls into the telescope T. Beam 2, having cleared the plate P, is reflected from the mirror 3₂, then reflected from the translucent side of the plate P and also falls into the tube T. (For clarity, rays 1 and 2 are depicted in different colors.) Ray 2 passes through the plate P three times, and the ray 1 - only once. To compensate for the difference in the path of the rays that arises because of this, a different plate P₁ is placed on the path of the first ray, made of the same materials as the plate P. The light beams 1′ and 2′ are correct and give an interference pattern in the field of view of the tube T. Depending on the difference in the path of the rays, the field of view of the pipe is light or dark. When the mirror 3₁ is moved by a distance equal to a quarter of the wavelength, the difference in the path of both rays increases by half the wavelength and the illumination of the visual field changes. Instead of a pipe, you can install a photomultiplier, which will produce electrical impulses at each level of illumination. This way you can count the number of waves that have decreased at a given length.

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Fig. 8. This photograph of viruses was taken using an electron microcoppe. The “large” sphere has dimensions of 2·10⁻⁷ m (2000Å). It is included for comparison.

LIST THE DISTANCE DO OF THIS GALAXY. It is now believed that the distance to some galaxies is approximately 10²⁶ m.

Let us now see how small extensions are determined.

It is not difficult to divide a meter into a thousand parts. It is a little more difficult to divide a millimeter into a thousand parts. For this you only need a good microcopper. This is how we get a micron - a millionth of a meter. But further division is difficult, since it is impossible to see objects shorter than the wavelength of visible light (about 5·10⁻⁷ m).

With the help of an electron microcoppe, we can see and change objects with dimensions up to 10⁻⁸ m. The shorter the wavelength of electron radiation, the smaller objects we can “see”. For example, gamma rays allow you to “examine” objects whose dimensions do not exceed 10⁻¹¹ m.

To determine nuclear dimensions, completely different methods are used: the so-called effective transverse section is measured.

reading of nuclei. It can be determined by passing a beam of high-energy particles through a thin plate of matter and measuring the number of particles that do not pass through. The ratio of the number of particles that did not pass through to all those emitted is proportional to the ratio of the area occupied by atomic nuclei to the area of ​​the plate. Similar experiments have shown that the radius of the nuclei ranges from 1·10⁻¹⁵ to 6·10⁻¹⁵ m*).

Finally, let’s talk about figure 7, which shows all the dimensional diversity of the world around us. “When solving scientific problems, a scientist always has to clearly imagine quantities in his imagination… those physical quantities that serve to describe the phenomenon being studied… Therefore, it is necessary to teach scientists from a young age so that the symbols in formulas that define physical quantities always represent specific, quantitative values ​​for them. For a physicist, unlike a mathematician, both parameters and variable quantities in a mathematical equation must be specific quantities,” says academician P. L. Kapitsa.

The scale unit figure 7 is meter. Each two adjacent divisions indicate sizes, perfectly exhausted from each other by 10 times. 10° m - one meter, 10¹ m - ten meters, 10² m - one hundred meters, etc. Similarly, 10⁻¹ m is a tenth of a meter (or ten centimeters), 10⁻² m is a hundredth of a meter (one centimeter), etc. Using logarithmic scales, you can show on one graph all sizes found in farming and technology: from the smallest to the largest.

*) A unit of length equal to 10⁻¹⁵ m is called a Fermi, including the famous physicist Enrico Fermi (1901-1954).

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RESULTS OF THE FINAL ROUND
ALL-UNION OLYMPIAD FOR SCHOOLCHILDREN

Math winners

From April 24 to April 30, the final round of the All-Union Mathematical Olympiad was held in Simferopol, in which 660 winners of regional and resource Olympiads took part.

The winners among eighth-graders were: Yuri Kolmakov from the city of Rasskazov, Tambov region, Andrey Kogan - a student at school No. 55 in Penza, Sergei Kuksin - a student at school No. 22 in Kharkov; among the ninth-graders: Alexey Alexandrov - a student at the physics and mathematics boarding school at Leningrad State University, Dmitry Logachev - a student at schools No. 2 in Moscow; among the tenth graders: Andrey Klimov and Andrey Khodulev - students of the physics and mathematics boarding school at Moscow State University.

31 participants of the Olympiad were awarded the second prize, 37 - the third, another 140 were awarded diplomas.

Physics winners

From April 9 to April 15, the final round of the All-Union Phys.

skoy Olympics. 652 winners of regional and resources Olympiads took part in it.

The competition took place in two stages. On April 10 there was a theoretical round, and on April 13 there was an experimental round.

The first places were taken by: eighth-graders Valery Marinich, a student at schools No. 1 in Dolinsk, Sakhalin Region, and Sergei Akhulkov, a student at schools No. 7 in Smolensk; ninth graders Igor Artyukhov - a student at the physics and mathematics boarding school at Moscow State University, Dmitry Kazakovtsev - a student at school No. 145 in Kyiv, Vladimir Nazaikinsky and Vladimir Fleishaker - both from schools No. 2 in Moscow; ninth-graders Sergei Gorbachevsky and Alexey Lomakin are students of the physics and mathematics boarding school at Leningrad State University.

16 Olympiad participants took second place, 40 took third place, 82 were awarded certificates.

The magazine «Kvant» congratulates all the winners of the Olympiad and wishes them further success.

A more detailed report will be published in one of the next issues of the journal.

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ru $V_B V_A$. Essentially we just need to check the vector from point $V_B$ to point $V_A$.

It is clear that the speed of any body $C$ on our plane must correspond to a certain point $V_c$, which is the end of the vector $\vec{V}_o V_c$ - the speed of body $C$ relative to body $O$. A vector drawn from point $V_c$ to any other point in the plane, for example, at $V_A$, is represented by the velocity vector of body $A$ relative to body $C$.

This allows us to represent the velocities of bodies not only as vectors, but also as points on a plane in velocity space. Any point on this plane means the speed of the body, and the vector going from one point to another is the vector of the relative speed of the body. It is clear what the meaning of the point $V_o$ with which we started is. This is the speed of the body $O$.

Of course, using velocity space, we do not distinguish between bodies located at different points in coordinate space, but having the same velocities. All of them correspond to the same point in velocity space. Therefore, by figure 4 we cannot say whether the body $C$ will reach the body $O$ or move away from it. This is determined generally by the condition of the problem that we are solving and by the positions of the bodies in coordinate space.

KINEMATIC DIAGRAM

Let us now move on to solving the scattering problem. Let a particle with mass $m_1$ and speed $v_1$ collide with another particle having mass $m_2$ and speed $v_2$, and let after the collision the velocities of the corresponding particles be $v_1'$ and $v_2'$). The task is to find the velocities

*) We believe that the particles move like billiard balls: the impact itself lasts a very short time, and the rest of the time the particles move as if they were free. Then the expressions “before the event” and “after the event” have a completely understandable meaning.

$v_1'$ and $v_2'$, if the vectors $v_1$ and $v_2$ and the particle masses are known.

The velocities $v_1'$ and $v_2'$ cannot, of course, have arbitrary values. Firstly, they must satisfy the law of conservation of momentum (momentum):

$$m_1 v_1' + m_2 v_2' = m_1 v_1' + m_2 v_2' ,$$

(1) secondly, they must satisfy the law of conservation of energy:

$$\frac{m_1 v_1^2}{2} + \frac{m_2 v_2^2}{2} = \frac{m_1 v_1'^2}{2} + \frac{m_2 v_2'^2}{2} .$$

In the case when the particles before and after the impact move in one straight line (such an impact is called central), these equations are solved quite simply. Instead of the vector equation (1), we can write exactly the same scalar equation for the lengths of the vectors. Then we get two scalar equations with two unknowns: $v_1'$ and $v_2'$. But more often you have to deal with an off-center impact, in which particles are separated at different angles. Having projected the particle velocity vectors into two mutually perpendicular directions, one of which is conveniently chosen to be parallel to the velocity of any of the particles before scattering, together with equation (1) we can write two exact scalar equations for the vector projections. Therefore, the conservation laws in the case of an off-center impact should be useful for the particles after scattering and the two angles that these vectors make with the direction we have chosen. If one of the quantities is specified, then the equation can be solved.

The collided part in any reporting system. Let’s take, for example, one in which particle 1 rummaged around before impact (such a system is called a laboratory system). It contains $v_1 = 0$, and the law of conservation of momentum looks like this:

$$m_2 v_2 = m_1 v_1' + m_2 v_2' .$$

This means that three vectors

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the number of $b$-kopeck coins and some other remainder, say $r$ kopecks. In this case, $r$ will certainly be less than $b$ (it is not excluded, of course, that $r = 0$).

The result of such an exchange can be written as follows:

$$a = kb + r, \quad r < b.$$

In this formula, the numbers $a, b$ are natural, that is, positive integers, and $k$ and $r$ are non-negative integers.

On the limit of division with a remainder. Dividing the integer $a$ by the natural number $b$ means representing $a$ as

$$a = kb + r,$$

where $k, r$ are integers, $b > r \geq 0$.

In this case, the number $k$ is called the quotient, and $r$ is the remainder of the division of $a$ by $b$.

In practice, division with a remainder is performed in the usual way, that is, “division by an angle.” For example,

$$\frac{175}{14} = \frac{14}{12};$$

$$\frac{28}{7}$$

here immediately there is a remainder and a quotient: $175 = 12 \cdot 14 + 7$.

Note that in this definition we do not require that $a$ be less than $b$, and we do not even require that $a$ be positive. You can, for example, divide

a) 5 by 7: $5 = 0 \cdot 7 + 5 \quad (k = 0, r = 5)$,

b) -13 to 7: $-13 = (-2) \cdot 7 + 1 \quad (k = -2, r = 1)$,

b) -224 by 7: $-224 = (-32) \cdot 7 \quad (k = -32, r = 0)$.

In the case where the remainder is zero, that is, $a = kb$, saying that $a$ is divisible by $b$.

It is clear that if the numbers $a$ and $b$ are divisible by the number $c$, then $a+b$ and $a-b$, as well as $ka, kb$ (k - natural) are divisible by $c$.

2. COMPARISON OF INTEGERS BY MODULE

Often there are problems in which the main role is played not by the numbers themselves included in the problem statement, but by the remainders from the division of these numbers.

Example one. Psesid goes from city $A$ to city $B$ 78 hours. He leaves $A$ at 10 o’clock in the morning. What time will he leave at $B$?

Remember that there are 24 hours in a day, divide the number 78 by 24:

$$78 = 3 \cdot 24 + 6$$

This means it will take three days and 6 hours to travel from $A$ to $B$. Therefore, he will arrive at $B$ at 16:00.

Please note that exactly the same result would have been obtained if it took not 78, but 54, 30 or 126, 102 and generally $24 \cdot k + 6$ hours to travel from $A$ to $B$ for any integer $k$.

Example two. A mathematics teacher conducts tests in class using the following system, which, in his opinion, ensures the need for mutual assistance among students. He gives them a piece of paper in a row with the conditions of the problem (“variants”), alters them according to the number of students in the class, and informs them that all the options are different. Indeed, students sitting close to each other solve various problems.

In fact, the teacher prepares only 7 different options: option 8 coincides with option 1, 9 - with 2, …, 14 - with 7, 15 - with 1, etc.

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The students soon figured out the teacher’s system, and at the next test, friends Ivanov and Petrov, sitting at opposite ends of the class, agreed to solve the same problems. The strong student Ivanov showed Petrov the number of his option with gestures, and Petrov exchanged options with a student sitting close to him, accurately establishing which option number he should have in order to receive a thank you cheat sheet from Ivanov.

You are probably reporting that Petrov figured out the numbers of the options - Ivanov’s and his own - by 7 and was only interested in the remainders of these divisions: he had to fill in the option with a number that gives the same remainder as the remainder of Ivanov’s option. The option numbers themselves were not significant—only the remainders were important.

For example, if Ivanov received option 10 (or 3, or 17), and Petrov received 26 (or 33, or 40), then dividing 10 and 26 by 7:

$$10 = 1 \cdot 7 + 3, \quad 26 = 3 \cdot 7 + 5,$$

Petrov exchanged options with a student who had an option

$$24 = 3 \cdot 7 + 3.$$

Example three. Let’s find out which digit the number $2^{999}$ ends up on. Consider successive powers of two:

$$2^1 = 2, \quad 2^2 = 4, \quad 2^3 = 8, \quad 2^4 = 16, \quad 2^5 = 32, \quad 2^6 = 64, \quad 2^7 = 128, \quad 2^8 = 256, \dots$$

It is easy to see that the last digits of these numbers are repeated after 4.

This means that the last digit of the number $2^n$ depends only on what remainder $n$ gives when divided by 4:

when $n$ is divisible by 4, that is, when $n = 4k$, all numbers of the form $2^n$ turn out to be 6,

at $n = 4k+1$ — » » 2,

at $n = 4k+2$ — » » 4,

at $n = 4k+3$ — » » 8.

The remainder of 999 divided by 4 is 3, so $2^{999}$ is 8.

Since solving this example it turned out that we only need to know what remainder $n$ gives when divided by 4, the possibility of all numbers was divided into four classes consisting of numbers $n$ of the form

$$4k, 4k+1, \quad 4k+2 \quad \text{и} \quad 4k+3.$$

And in the second example (with test options), the possibility of all options was divided into 7 classes of the form

$$7k, \quad 7k+1, \quad 7k+2, \dots, \quad 7k+6.$$

In general, if we are interested in remainders when divided by some fixed number $m$, then the set of all integers (not only positive ones!) is divided into $m$ classes. Each class consists of numbers that give identical remainders when divided by $m$. These are the classes:

0. numbers $a$ of the form $a = km,$

1. » $a$ » $a = km+1,$

2. » $a$ » $a = km+2,$

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… … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

30

Let us now introduce the following definition:

Defined. If the integers $a$ and $b$ it is possible that their difference is divisible by the natural number $m$, then saying that $a$ and $b$ are comparable modulo $m^*$).

Let’s compare the numbers $a$ and $b$ modulo $m$ written like this:

$$a \equiv b \pmod{m}.$$

It is clear that two numbers are comparable modulo $m$ if and only if they give identical remainders when divided by $m$, that is, they belong to the same class.

For example,

$$27 \equiv 7 \pmod{10}, \quad 78 \equiv 6 \pmod{24}, \quad 6 \equiv 0 \pmod{3}, \quad 25 \equiv -1 \pmod{13}.$$

Clear comparability of numbers turns out to be useful in solving many problems.

Problem 1. For what natural numbers $n$ is the number $n^2+2$ divisible by 3, that is, $n^2+2 \equiv 0 \pmod{3}$?

Solution.

If $n \equiv 0 \pmod{3}$, then $n^2+2 \equiv 2 \pmod{3}$,

» $n \equiv 1 \pmod{3}$, » $n^2+2 \equiv 0 \pmod{3}$,

» $n \equiv 2 \pmod{3}$, » $n^2+2 \equiv 0 \pmod{3}$.

So, $n^2+2 \equiv 0 \pmod{3}$ if and only if $n$ is not divisible by 3.

The basic properties of comparison are reminiscent of the properties of general equalities. It is comparisons that can be obtained by adding, subtracting and multiplying:

if $a \equiv b \pmod{m}$, and $c \equiv d \pmod{m}$, then

$$a + c \equiv b + d \pmod{m}, \quad a - c \equiv b - d \pmod{m}, \quad ac \equiv bd \pmod{m}.$$

Let us prove, for example, that comparisons can be multiplied. Since $a \equiv b \pmod{m}$ and $c \equiv d \pmod{m}$, then $a-b$ and $c-d$ are divisible by $m$. From equalities

$$ac - bd = ac - ad + ad - bd = a(c-d) + d(a-b)$$

it is clear that $ac-bd$ is also divisible by $m$, that is

$$ac \equiv bd \pmod{m}.$$

As an exercise, prove that the comparison can be added and subtracted.

Let $a \equiv b \pmod{m}$. From the above properties of comparison it follows that $a^k \equiv b^k \pmod{m}$ for any natural number $k$ (that is, that both sides of the comparison can be raised to a power) and, in addition, that $ac \equiv bc \pmod{m}$, where $c$ is any integer (that is, that both sides of the comparison can be multiplied by any integer).

Problem 2. What remainder when divided by 3 gives the number $(2^2+1)(3^2+1)(4^2+1) \dots (1000^2+1)$?

Solution. This remainder will not change if each of the numbers is 2, 3, . . . , 1000 in this expression, replace it with the remainder when divided by 3. Thus, the number of interest modulo 3 is equal to $(2^2+1)^{333} (3^2+1)^{333} (1^2+1)^{333} \equiv 5^{333} \cdot 10^{333} \cdot 2^{333} \equiv 2^{333} \cdot 1^{333} \cdot 2^{333} \equiv 2^{2 \cdot 333} \equiv 4^{333} \equiv 1^{333} \equiv 1$, i.e., the required remainder is 1.

*) The word “module” (from the Latin “modulus” - measure) is used in various areas of mathematics and its applications; it is assigned to a number of particular importance; for example, counting or measuring is carried out in relation to it.

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Task 3. Prove that every natural number $n$ is comparable modulo 9 to its sum of digits.

Solution. Let

$$\overline{a = a_n a_{n-1} \dots a_0},$$

where $a_0, a_1, \dots, a_n$ are the digits of the number $a$. We have

$$a = a_n \cdot 10^n + a_{n-1} \cdot 10^{n-1} + \dots + a_0.$$

Let us now write down the following estimated comparisons:

$$1 \equiv 1 \pmod{9}, \quad 10 \equiv 1 \pmod{9}, \quad \dots, \quad 10^n \equiv 1 \pmod{9}.$$

Multiplying these comparisons by $a_0, a_1, \dots, a_n$ respectively and adding, we get

$$a_0 + 10 \cdot a_1 + \dots + 10^n a_n = a \equiv (a_0 + a_1 + \dots + a_n) \pmod{9}.$$

From here, by the way, we immediately get the well-known sign of divisibility of numbers by 9 (and 3).

3. ARITHMETICS OF REMAINS ACCORDING TO THIS MODULE

Let a natural number $m$ (modulo) be given. Considering all possible integers, we will be interested only in their remainders when dividing them by $m$. Thus, we will not distinguish numbers comparable modulo $m$ from each other. Under this condition, we have a remainder of $m$ of “different” objects - remainders from division by $m$:

$$0, 1, 2, 3, \dots, m-1,$$

and for them we can define new rules for “addition” and “multiplication”: namely, if $r_1$ and $r_2$ are two remainders, then “sum” $r_1 + r_2$ we will call the remainder that gives the number $r_1 + r_2$ when divided by $m$, and “product” $r_1 \cdot r_2$ is the remainder when dividing $r_1 \cdot r_2$ to $m$. For example, for $m = 10$ the remainder when divided by 10 is the last digit of its decimal notation. Therefore, in arithmetics of remainders modulo 10 will be true, for example, also the equalities

$$7 + 3 = 0, \quad 1 + 2 = 3, \quad 7 + 5 = 2, \quad 7 \cdot 5 = 5, \quad 7 \cdot 6 = 2, \quad 2 \cdot 5 = 0.$$

In the .ariphmetic of modulo 2 remainders there are only two remainders: 0 and 1, so this .ariphmetic is quite simple:

$$0 + 0 = 0, \quad 1 + 0 = 1, \quad 1 + 1 = 0,$$

$$0 \cdot 0 = 0, \quad 1 \cdot 0 = 0, \quad 1 \cdot 1 = 1.$$

For $m = 6$ we get

$$5 + 3 = 2, \quad 4 + 2 = 0, \quad 3 + 3 = 0, \quad 3 \cdot 2 = 0, \quad 3 \cdot 4 = 0$$, etc.

Thus, whichever modulo $m$ corresponds to its own ariphmetic. For the sake of brevity, we will call it $m$-ariphmetic*).

32

As an example, we give multiplication and addition tables in 7-arithmetic.

Multiplication table

| × | 1 | 2 | 3 | 4 | 5 | 6 |
| :— | :— | :— | :— | :— | :— | :— |
| 1 | 1 | 2 | 3 | 4 | 5 | 6 |
| 2 | 2 | 4 | 6 | 1 | 3 | 5 |
| 3 | 3 | 6 | 2 | 5 | 1 | 4 |
| 4 | 4 | 1 | 5 | 2 | 6 | 3 |
| 5 | 5 | 3 | 1 | 6 | 4 | 2 |
| 6 | 6 | 5 | 4 | 3 | 2 | 1 |

Addition table

| + | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| :— | :— | :— | :— | :— | :— | :— | :— |
| 0 | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| 1 | 1 | 2 | 3 | 4 | 5 | 6 | 0 |
| 2 | 2 | 3 | 4 | 5 | 6 | 0 | 1 |
| 3 | 3 | 4 | 5 | 6 | 0 | 1 | 2 |
| 4 | 4 | 5 | 6 | 0 | 1 | 2 | 3 |
| 5 | 5 | 6 | 0 | 1 | 2 | 3 | 4 |
| 6 | 6 | 0 | 1 | 2 | 3 | 4 | 5 |

These tables are devices like this: at the intersection of rows and columns, they build the product (correspondence - sum) of the remainders at the beginning of these rows and columns.

From the properties of comparison and the usual rules of action on numbers it is clear that in $m$-arithmetic the given rules of ordinary arithmetic are satisfied, that is, for any remainders $a, b, c$ there will be

$$a+b = b+a \text{ — коммутиативность (переместительность) сложения},$$
$$(a+b)+c = a+(b+c) \text{ — ассоциативность (сочетательность) сложения},$$
$$ab = ba \text{ — коммутиативность умножения},$$
$$(ab)c = a(bc) \text{ — сочетательность умножения},$$
$$a(b+c) = ab+ac \text{ — дистрибутивность (распределительность) умножения относительно сложения}.$$

Thus, in our arithmetics, for example, the following formulas will be correct:

$$(a+b)^2 = a^2 + 2ab + b^2,$$
$$(a+b)(c+d) = ac + ad + bc + bd.$$

Let’s consider several problems using $m$-arithmetic.

Task 4. Find the remainder when $2^{1000}$ is divided by 7.

Solution. We need to figure out what $2^{1000}$ is equal to in 7-arithmetic. It is easy to see that $2^{1000} = (2^3)^{333} \cdot 2 = 1^{333} \cdot 2 = 2$.

Task 5. Find the last digit of the number $7^{7^{1000}}$.

Solution. We have to figure out what this number is equal to in 10-arithmetic. To do this, consider the successive powers of the number 7 in 10-arithmetic:

$$7^1 = 7, 7^2 = 9, 7^3 = 3, 7^4 = 1, 7^5 = 7 \text{ и т. д.}$$

It’s clear that

$$7^{4n} = 1, 7^{4n+1} = 7, 7^{4n+2} = 9, 7^{4n+3} = 3.$$

So, we need to find out the remainder when dividing the number $7^{1000}$ by 4. To do this, consider $7^{1000}$ in 4-arithmetic. Since $7^{1000} = 3^{1000}$, we now consider the sequence of remainders in 4-arithmetic:

$$3^1 = 3, 3^2 = 1, 3^3 = 3, 3^4 = 1, 3^5 = 3, \dots,$$

33

$$3^{2n+1} = 3, \quad 3^{2n} = 1.$$

But 1000 is an even number; means $7^{1000} = 1$, that is $7^{1000} = 4n+1$. Therefore, in 10-arithmetic $7^{71000} = 7^{4n+1} = 7^{4n} \cdot 7 = 7$.

Answer: the number $7^{71000}$ ends in 7.

Problem 6. Prove that if $a^2+b^2$ is divisible by 3, then both $a$ and $b$ are divisible by 3.

Solution. Passing to remainders from division by 3, we obtain the 3-arithmetic equality $a^2+b^2 = 0$. There are three remainders in 3-arithmetic: 0, 1, 2.

We have: $1^2 = 1, 2^2 = 1$. It is easy to see that for $a \neq 0$ $a^2+b^2 \neq 0$, that is, from the equality $a^2+b^2 = 0$ it follows $a = 0$ and $b = 0$. This means that $a$ and $b$ are simultaneously divisible by 3.

TASKS *)

1. Division with remainder
   1.0. Divide with the remainder -297 by 31, -1005 by 98.

2. Prove that the number of divisors of any natural number $N$ does not exceed $2\sqrt{N}$.

3. Find remainders from division of numbers ($n$ - natural): a) $n$ by $(n-1)$ and by $(n-2)$; b) $(n^2+n+1)$ on $(n+1)$ and on $(n+2)$; c) $(n^4+1)$ to $(n+3)$ (at $n \geqslant 80$).

4. Find all integers $n$ for which $\frac{n^2+1}{n-1}$ is also an integer.

5. Prove that the product of any three consecutive integers is divisible by 6.

6. Prove that if $a+b+c$ is divisible by 6, then $a^3+b^3+c^3$ is also divisible by 6 ($a, b, c$ are integers).
   7*. 15 prime numbers form an arithmetic progression. Prove that its difference is greater than 30,000.

7. Comparing integers modulo
   8.0. Find all natural numbers $n$ for which the number $n^3+n+1$ is divisible by 3.
   9.0. Solve the comparison $6n+5 \equiv 0$ (mod 7).

8. Prove the following test for divisibility by 11:
   $$a = \overline{a_n a_{n-1} \dots a_0} \equiv 0 \pmod{11}$$

then and only when
$(-1)^n a_n + (-1)^{n-1} a_{n-1} + \dots + a_n \equiv 0 \pmod{11}$.
11. Using the fact that $1000 \equiv -1$ (mod 7), as well as $1000 \equiv 1$ (mod 37), obtain divisibility criteria for 7 and 37.
12. Prove that if the numbers $A$ and 5$A$ have the same sum of digits, then $A$ is divisible by 9.
13. Prove that the equation $15x^2-7y^2=9$ has no integer solution.
14. Prove that for no integer $n$ the number $n^2+3n+5$ is divisible by 11.
15. Prove that numbers of the form $3n+2, 5n \pm 2, 7n+3, 7n-1, 7n-2$ are not perfect squares for any natural number $n$.
3. Arithmetic of remainders for a given modulus
16.0. Construct addition and multiplication tables in 4-, 6-, 8-, 11-, and 13-arithmetic.
17. Construct tables of squares and cubes in 9-arithmetic.
18. Using the table of cubes obtained in Problem 17, prove that numbers of the form $9n+5$ and $9n+4$ cannot be represented as the sum of the cubes of three numbers.
19. Prove that in $p$-arithmetic, where $p$ is a prime number, from the equality $ab=0$ with $a \neq 0$ it follows that $b=0$.
20. Prove that in $m$-arithmetic for any remainder $a$ there is a unique remainder - $a$ such that $a+(-a)=0$. Deduce from this that the equation $a+x=b$ has a solution in any $m$-arithmetic for any $a$ and $b$.
21. Prove that for composite $m$ in $m$-arithmetic there are remainders $a \neq 0$ and $b \neq 0$ as well as $ab=0$.
22. Using 31 arithmetic, prove that $1^{255}+2^{255}+\dots+30^{255}$ is divisible by 31.

*) Easy tasks are marked with a zero, and the most difficult ones are marked with an asterisk.

ALBERT EINSTEIN

ELEMENTARY
THEORY
FLIGHT
AND WAVES
ON THE WATER

Everyone now knows why an airplane can stay in the air, but sixty years ago it was…
Pols, a riddle. The more interesting

35

watch how Albert Einstein, the great scientist, creator of the theory of relative explanation, discovered the lifting force of the wings of airplanes and birds.

The published article is taken from the collection of scientific works of A. Einstein, published by Nauka in 1967 (vol. 4, p. 22). Einstein wrote his work in 1916. He was not familiar with the works of N. E. Zhukovsky, who back in 1910 created theories of the flight of heavier-than-air vehicles.

Where does the lifting force of the wings of our airplanes and birds soaring in the air come from? There is complete uncertainty in these matters. I must admit that I could not find even the simplest answer to them in the specialized literature. I hope, therefore, that the reader will be pleased enough if I tried to take advantage of this gap with the help of the following simple considerations from the theory of fluid motion.

The incompressible fluid, whose internal friction we will neglect, flows along the existing pipe (Fig. 1) in the direction indicated by the arrows. We will be interested in the pressure distribution in the pipe. Since the same amount of liquid must flow through each section per unit time, the flow rate $v$ will be greatest where the cross-sectional area is minimal, and least where the cross-sectional area is maximum. Therefore, at figure 1, ​​the velocity of part of the fluid is the smallest at point $L$ and continuously increases towards $R$. The reason that causes such acceleration of a part of the liquid is nothing more than the pressure force acting on them. Let us consider a particle $F$ of a liquid occupying a cylindrical volume. In order for a fluid particle $F$ to have an acceleration directed to the right at a given moment, the pressure on its rear surface $A$ must be greater than the pressure on its front surface $B$. The pressure on surface $A$ exceeds the pressure on surface $B$. Repeating these arguments, we come to the conclusion that the pressure in the pipe continuously drops from $L$ to $R$. We will obtain the same pressure distribution (pressure decrease from $L$ to $R$) using similar reasoning in the case when the direction of fluid flow changes to the opposite.

Summarizing what has been said, we can formulate the following well-known theorem of hydrodynamics of a non-appearing fluid. If we follow the trajectory of any fluid particle in a steady flow, then the pressure $p$ will always be greater where its speed $v$ is less, and vice versa. As is known, the number of expressions of this theorem for

36

non-removable liquid has the form

$$p = \text{const} - \frac{1}{2} \rho v^2,$$

where $\rho$ is the density of the liquid.

Let us first consider one well-known example illustrating this theorem—the flow of a liquid under constant pressure from a hole (Torricelli). At point $J$ (Fig. 2) the pressure is greater, and the speed, on the contrary, is less than at point $A$, so the expression

$$p + \frac{1}{2} \rho v^2$$

constantly in the stream.

As a second example, consider the spray gun (Fig. 3). The air flow passing through the tube $L$, after leaving the hole, expands in all directions, while reducing its speed. Therefore, the pressure at point $P$ is less than at point $G$, and, therefore, less than in the air at rest surrounding point $P$. Due to the applied pressure at point $P$, the liquid from the vessel $A$ rises upward and is sprayed into small particles by the air flow

droplets. (The fact that in this example we are dealing with a flow of air, and not with a flow of undischarged liquid, does not essentially change anything in our reasoning.)

After these preparations, let us turn to the consideration of waves on water. Let $W$ be a solid wall in the form of a wavy cylinder and located perpendicular to the drawing plane, on one side bordering on a fluid flow flowing from left to right (Fig. 4). We will be interested in the force with which the liquid acts on the wall. It is clear that the cross-section of the fluid flow at points $B$ is larger than at points $T$. Consequently, near points $B$ the liquid will flow slower, and near points $T$ - faster than at those points inside the liquid that are located away from the wall $W$. Therefore, near points $B$ the fluid flow will create excess pressure, and near points $T$ a vacuum will be observed. As a result, the liquid will press on the wall as if it were trying to increase its bend. This means that the flow could not be sustained,

37

if the surface of the liquid were free and, accordingly, if the wall could bend and stretch without limit *).

In these discussions, as before, we proceeded from the assumption that there are no causes causing pressure other than the flow of fluid. If the force of gravity acts in the direction of the arrow S, then it leads to the appearance of pressure in the liquid, returning from top to bottom. If only one gravity acted, then the pressure at points B would be less than at points T.

So, flow and gravity give rise to the prerequisites for the appearance of various pressure differences between points B and T. It is clear that it is possible to select the speed of fluid flow in such a way that the resulting exchange pressure differences between points B and T will be equal to zero. After this, wall W can be removed without introducing any disturbances during

*) It is known that these same messages can explain why the flag will lie in the wind.

Fig. 4

Fig. 5

liquids. As a result, we get a fluid flow with a wave-like excluded surface, which part can be observed when the flow flows around some obstacle. We see this picture when looking from a place into the water if the value is above the support.

If we decide to describe the whole process from the point of view of an observer moving to the right with the speed of the flow away from the walls, then we will come to ordinary waves on the surface of the water. To do this, it is observed at rest, and the crests B and troughs T float back at a constant speed.

Consequently, the possibility of wave-forming processes is based on the fact that static and dynamic pressure differences that arise between points with different heights cancel each other out.

The explanation for the reasons for the appearance of the lifting force of the wing looks completely similar. Let a solid wall be inserted into a flow of liquid or air, located parallel to the

38

ku and perpendicular to the plane of the drawing (Fig. 5), on the upper surface of which it looks like a plane. If this convexity did not exist, then no forces would act on the surface of the wall, except for the inevitable friction. The convexity will affect the flow of fluid at both the upper and lower surfaces of the wall, and thus creates additional pressure.

For a flow flowing around a bluish wall, the bulge creates a local increase in the cross section and, consequently, a slowdown in the flow; as a result of this, the pressure at point U will increase. On the upper surface, on the contrary, convexity means a decrease in the cross section, which means a local increase in the flow velocity and thereby a drop in pressure at point O. Thus, the dynamic forces of pressure, arbitrary by the flow, create a force acting on the wall and directed upward. It is clear that for the appearance of this force it is only necessary that the piece of the wall be as large as required for noticeable bending of the fluid flow. We get a non-existent wing of an airplane or a bird (not flapping its wings in flight).

Already from these simplest arguments it is clear that flight requires only a certain power, since it is necessary to overcome the resistance of inevitable friction. If there were no friction, birds could fly any horizontal distance without expending any work.

TASK

Census Agent Smith and Survey Agent Jones simultaneously approach house number 900. Each owner indicates the age of the animal in the house. The owner of the house (produced in the USA) tells them his age and says that there are three other residents living in the house, whose ages are three different numbers - when changing days, the number is equal to the house number. The homeowner says he will tell the census agent the ages of the residents. He whispers this age to the census agent, who then says that he is unable to determine the ages of the other two residents. Then the House Owner says that he will tell the survey agent the sum of the ages of the eldest of the occupants and one of the other two. He whispers the amount to the survey agent, who says that he too is unable to determine the animal’s age.

The owner of the house begins to ask them one by one. The first time the census agent answers is that he can determine these ages. The polling agent says he can determine the ages. The survey agent says he still doesn’t know the ages of the residents, and the survey agent states, “I know all the ages now.” What are the ages of these three residents? (The problem statement contains all the necessary information for the solution!)

PROBLEM WORKER
quantum

F24. stand,
on which
the body lies
suspended
outside,
start
let down
acceleration α.
In the initial
moment
spring
not stretched.
Through
what time is it
the body will come off
from the stand?
Do what
maximum
length
will the spring stretch?
Body weight M,
spring stiffness κ.
N. I. Goldfarb

39

40

F25. If the potential of the anode of the photozement is higher than the potential of the cadada, then a current $I=10 \, a$ (so saturation) flows through the photozement. Otherwise, no current flows through the photozesment. Neglecting the internal resistance of the batteries, find the voltage across the photozems in the diagrams shown in figure 1 (resistance values ​​are indicated in kilums).

F26. Two horizontal half-planes, located at a height $h$ one above the other, smoothly transform into each other, as shown in figure 2. Along the upper half-plane at an angle $\alpha$ to the direction of descent, a small block moves at a speed $U$. How will it move along the lower half-plane? Assume that the block does not bounce, that is, it moves without leaving the surface of the descent. Neglect training.

F27. Water is poured into a hemispherical bell, lying tightly on the table, through the hole at the top (Fig. 3). When the water reaches the hole, it lifts the bell and begins to flow out. Find the weight of the bell if its radius is $R$ and the density of water is $\rho$.

42

Modern industry cannot do without a wide variety of crystals. They are used in watches, transistor receivers, computers, lasers and many others. The great laboratorium - nature - can no longer satisfy the demand of developing technology, and so artificial crystals are grown in special factories: small, almost imperceptible, and large ones - weighing several kilograms.

There are different ways to grow crystals. Often this process requires high temperatures and enormous pressure (for example, to obtain artificial diamonds), but some crystals can be grown even at home. We will tell you how this can be done.

The easiest way to grow crystals of aluminum alum at home is KAl(SO₄)₂·12H₂O. This substance can be bought at any chemical store, and it is absolutely harmless (alum is sometimes even added to drinking water to clear it of turbidity). But before we get started, let’s take a look at what the crystal growing process is all about.

If you dissolve a substance in water at a constant temperature, then after some time, what appears to be the dissolution process stops. Such a solution is called saturated, and the maximum amount of a substance that can be dissolved at a given temperature in 100 °C water is called its solubility. Obviously, with increasing temperature, solubility increases. Therefore, a saturated industry cannot do without a wide variety of crystals. They are used in watches, transistor receivers, computers, lasers and many others. The great laboratorium - nature - can no longer satisfy the demand of developing technology, and so artificial crystals are grown in special factories: small, almost imperceptible, and large ones - weighing several kilograms.

There are different ways to grow crystals. Often this process requires high temperatures and enormous pressure (for example, to obtain artificial diamonds), but some crystals can be grown even at home. We will tell you how this can be done.

The easiest way to grow crystals of aluminum alum at home is KAl(SO₄)₂·12H₂O. This substance can be bought at any chemical store, and it is absolutely harmless (alum is sometimes even added to drinking water to clear it of turbidity). But before we get started, let’s look at what the dissolution process is like when it stops. Such a solution is called saturated, and the maximum amount of a substance that can be dissolved at a given temperature in 100 °C water is called its solubility. Obviously, with increasing temperature, solubility increases. Therefore, a saturated industry cannot do without a wide variety of crystals. They are found in watches, transistor receivers, computers, lasers and many others. The great laboratorium - nature - can no longer satisfy the evolving technology, and so artificial crystals are grown in special factories: small, almost imperceptible, and large ones - weighing several kilograms.

There are different ways to grow crystals. Often this process requires high temperatures and large amounts of water, which can cause damage to the crystal structure. Therefore, it is recommended to use a specialized laboratory for growing crystals.

43

saturated at one temperature, becomes undersaturated at a higher temperature. If the saturated solution is cooled, the excess substance precipitates. figure 1 shows the dependence of the solubility of aluminum alum on temperature. The graph shows that, for example, when cooling to 10°C 100 g of a solution that was saturated at 30°C, more than 10 g of the substance should precipitate. Therefore, one way to grow crystals is to allow the saturated solution to cool.

Crystals can also be grown by evaporation. After all, if a saturated solution evaporates, its volume decreases, but the amount of dissolved substance remains the same. In other words, an excess of substance is again created, which precipitates.

Let us now consider how the release of excess substance occurs.

Take a saturated solution and heat it. Cover the vessel with the resulting undersaturated solution with glass and allow the solution to cool quietly to a temperature lower than the saturation temperature. In this case, the precipitate may not fall out, and we will get a supersaturated solution. The fact is that for education

To form a crystal, a “seed” is necessary. It can be a small crystal of the same substance or a simple speck of dust. Sometimes it is enough to quickly remove a vessel with a supersaturated solution or remove the glass covering it, as instant crystallization begins. In this case, many small crystals are usually formed. In order to grow a large crystal, it is necessary to limit the number of “seeds”. It is best to introduce an artificial “seed”, the role of which will be played by one of the crystals obtained earlier.

The “seed” is prepared as follows. Take two glass jars and wash them thoroughly. Pour warm water into one of them and saturate it with alum. When placing the solution, watch for dissolution. When the substance stops dissolving, carefully pour the solution into the second jar so that no undissolved substance gets into it. Then cover the jar with glass. When the solution has cooled, chemistry the glass. After a while, you will see many crystals forming in the jar. Let them grow and select the largest ones for “seeding”.

Now you can start growing the crystal. First of all, you need to prepare the dishes. To destroy unwanted items -

44

Shea on the walls, steam the inside of the jars over the spout of a boiling kettle. Then make a warm, saturated solution again and pour it into another clean jar.

So, you have a warm saturated alum solution. Heat it a little more, cover the jar with glass and leave to cool. When the temperature of the solution approaches the saturation temperature, put the previously prepared “dressing” into the jar. Since the solution is still undersaturated, the “priming” crystal begins to dissolve. But as soon as the solution cools to the saturation temperature, the dissolution of the crystal will stop, and soon its growth will begin. (If the crystal dissolves entirely, you can add a new “filling” to the solution.))

When the solution stops cooling, crystal growth can be continued. To do this, lift the glass so that the water evaporates, but dust particles do not fall into the solution. Crystal growth continues for two to three days.

When growing a crystal, be careful not to touch or move the jar. When the crystal is ready, just remove it from the solution and blot it thoroughly with a paper towel, otherwise it will quickly fade.

The crystals come out differently in shape depending on whether you drop the dressing at the bottom of the sauce or hang it on a string (Fig. 2). In this way you can, for example, grow “beads”. To do this, you need to “thread” the thread, that is, pass it several times along the crystal, and then lower the thread into the solution.

The photo of the graphic placed at the beginning of the stats shows a crystal obtained from potassium alum using the above method.

Growing crystals is an art. You may not succeed in everything right away. Don’t be upset. A little persistence, perseverance, accuracy, and you will become the owner of beautiful crystals.

To questions from readers

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Where is the logical fallacy?

In The Amer magazine. math. Monthly a task was placed, which in translation sounds like this: “from what point on the globe do you need to leave in order to walk 10 km along the meridian to the south, then 10 km along the parallel to the east, and finally, again 10 km along the meridian to the north, to arrive at the point of departure?”

One solution is almost very visible: exit the North Pole. Interestingly, this solution is not the only one. One prominent mathematician addressed this problem in this way. The first and third parts of the path pass along the meridians; but the two meridians have only two common points - the northern and southern stripes; the latter disappears, since it is impossible to move south from it; the northern lane remains as the only solution.

What is the reasoning for the error and how to get the complete solution?

45

ABOUT FOSTERS
EXAMINATIONS
IN MATHEMATICS
IN TECHNICAL
UNIVERSITIES IN 1969
V. A. TONYAN

EXAMINATIONS AT MIEM

In this article we will talk about the entrance exams in mathematics to the Moscow Institute of Electronic Engineering (MIEM).

As in most technical schools, applicants take two exams in mathematics: written and oral. 4 hours (4×60 minutes) are allotted for written work.

Please note that all proposed questions and tasks do not go beyond the approved program for entrance exams.

Below we present problems taken from the various versions of the written examinations offered in 1969, indicating the most common mistakes made by applicants.

Option 0

1. A motorcyclist and a bicycle rider simultaneously head towards each other from points A and B and meet at a distance of 12 km from the middle of AB. The speed of a motorcycle rider is 32 km/h higher than the speed of a bicycle rider. If the motorcycle rider had left an hour later, they would have met halfway along the way. Find the distance AB.

2. Solve the equation
   $$x = a - \sqrt{a^2 - x \sqrt{x^2 + a^2}}.$$

3. Solve inequality
   $$\log_{\sqrt{3}} \frac{\sqrt{27}}{\arctg x} \cdot \log_{\sqrt{3}} \arctg x > 2.$$

4. Solve the equation
   $$\sin \sqrt{4 + 3x - x^2} + \cos \sqrt{4 + 3x - x^2} = \sqrt{2}.$$

5. The centers of two circles, whose radii are R and r, lie on the hypotenuse of a right triangle. One circle touches two legs, the other touches the leg and the first circle. Find the sides of the triangle.

Let us present the solution to these problems.

1. Let’s denote the speed of the speedster by x, and the speed of the speedster AB by 2S; then the motorcyclist’s speed will be x+32. If the time spent by the motorcyclist before the meeting is equal to t, then from the conditions of the problem we obtain
   $$xt = S - 12,$$
   $$(x + 32)t = S + 12.$$

If 动机iklist had left an hour later, he would have spent $t_1$ hours before the meeting, while 动机iklist would have spent $(t_1-1)$ hours and therefore
$$(x + 32)t_1 = S,$$
$$x(t_1+1) = S.$$

Solving this system, we obtain $S=48$ km.

This task turned out to be easy for applicants. The overwhelming number of applicants solved it. However, a lot of cumbersome systems of equations are due to poor choice of notation.

2. By squaring both sides of this equation, we get
   $$x^2 - 2ax = -x \sqrt{x^2 + a^2}$$
   or
   $$x(2a - x - \sqrt{x^2 + a^2}) = 0.$$

From here
$$x_1 = 0$$ and $2a - x = \sqrt{x^2 + a^2}$.

Squared again, we get $3a^2 = 4ax$ or $a(3a-4x)=0$. Now if a=0,

46

then $x_2 - \text{любое число},$ if $a \neq 0$ then
$$x_2 = \frac{3}{4} a.$$

When performing the transformation (squaring), we could violate the equivalence, so the found roots $x_1, x_2, x_3$ must be checked.

1. $x_1 = 0$. We substitute the value $x_1$ into the original equation. Then the left side is 0, and the right side is:
   $$a - \sqrt{a^2} = a - |a| = \begin{cases} 0 & \text{при } a \geq 0, \\ 2a & \text{при } a < 0. \end{cases}$$

2. If $a = 0$, then the left side of the equation is $x$, and the right side is:
   $$- \sqrt{-x \sqrt{x^2}} = -\sqrt{-x |x|} = \begin{cases} -\sqrt{-x^2} & \text{при } x > 0, \\ -\sqrt{x^2} & \text{при } x \leq 0. \end{cases}$$

3. Substituting the value $x_3 = \frac{3}{4} a$, we find that the left side is equal to $\frac{3}{4} a$, and the right side is:
   $$a - \sqrt{a^2 - \frac{3}{4} a \sqrt{\frac{9}{16} a^2 + a^2}} = \begin{cases} a - \sqrt{\frac{a^2}{16}} = \frac{3}{4} a & \text{при } a > 0, \\ a - \sqrt{\frac{31 a^2}{16}} + a + \frac{a \sqrt{31}}{4} & \text{при } a < 0. \end{cases}$$

Answer. If $a > 0$, then $x_1 = 0$, $x_3 = \frac{3}{4} a$; if $a = 0$, then $x \leq 0$.

A far more typical mistake that applicants made when solving this equation was not checking the found values ​​of $x_1, x_2, x_3$. When solving this equation using the above method, verification is an integral part of solving the problem.

3. Denoting $\log \sqrt{3}$ arcctg $x = t$, we get $t^2 - 3t + 2 < 0$, hence $1 < t < 2$.
   Returning to the variable $x$, we can write
   $$\sqrt{3} < \text{arcctg} x < 3$$
   where we get the answer: $\text{arcctg} 3 < x < \text{arcctg} \sqrt{3}$.

4. Requiring $4 + 3x - x^2 \geq 0$ gives us $-1 \leq x \leq 4$.

Now, applying the addition formulas to the left side, we get
$$\sqrt{2} \sin \left(\sqrt{4 + 3x - x^2} + \frac{\pi}{4}\right) = \sqrt{2},$$
where
$$\sqrt{4 + 3x - x^2} = \frac{\pi}{4} + 2\pi n.$$
This means that $n \geq 0$ is necessary.

Solving this irrational equation, we get
$$x = \frac{3 \pm \sqrt{25 - 4 \left(2\pi n + \frac{\pi}{4}\right)^2}}{2};$$
from here
$$\left(2n\pi + \frac{\pi}{4}\right)^2 = \frac{25}{4}.$$
Therefore, $n = 0$ and $x_{1,2} = \frac{6 \pm \sqrt{100 - \pi^2}}{4}$.

By checking we can easily verify that both of these values ​​are roots of the equation.

In solving this equation, many obtained equality (1), but only part of the constructional studies of this formula left a solution corresponding to $n = 0$.

5. Let $ABC$ be a given triangle, $\rightarrow ACB$ a straight line, $O$ and $O_1$ the centers of circles of radii $R$ and $r$; $D$ and $E$ – points of tangency of the circles of leg $CA$. Let’s draw $EF \parallel AB$ and denote $\rightarrow BAC = \alpha$. Then from $\triangle FDE$ we have
   $$\sin \alpha = \frac{R - r}{R + r}.$$

Noticing that $CD = R$ and denoting $AC = b$, we obtain from $\triangle ODA$
$$\text{tg} \alpha = \frac{R}{b - R}.$$

From (1) and (2) we can write
$$1 + \left(\frac{R}{b - R}\right)^2 = \left(\frac{R + r}{R - r}\right)^2,$$
whence, solving for $b$, we obtain
$$b = R \left[1 + \frac{2\sqrt{Rr}}{R - r}\right].$$

47

---

running_head:
folio:

Now you can find the other side and the hypotenuse. (We suggest you check the calculation yourself.)

In conclusion, we present some options for written work offered at the entrance exams in 1969.

Option 1

1. Two tourists left the city at the same time in the same direction. The first tourist travels every kilometer of the journey 20 minutes faster than the second. Having walked part of the way, the first tourist turned back, and after walking 1 km, he went in the same direction. When the first tourist caught up with the second, the second one had been traveling for 1/2 hour. How many minutes does it take the first tourist to walk 1 km?

2. Solve the equation

$$\sin \sqrt{4 + 3x - x^2} + \cos \sqrt{4 + 3x - x^2} = \sqrt{2}.$$

3. Solve inequality

$$\frac{\log_2(x^2 - 3x - 4)}{\log_2(x + 17)} \leq 1.$$

4. Solve inequality

$$\sin^2 2x (\text{ctg} x - \text{tg} x) > 2 \cos 2x.$$

5. The lateral edges of the triangular paramid are mutually perpendicular. The height of the paramid is $h$. Find the radius of the sphere circumscribed about the paramid.

Option 2

1. Two diggers were tasked with digging a ditch in 3 hours and 36 minutes. However, the first one started work when the second one had already dug half of it and stopped digging. As a result, the ditch was dug in 7.5 hours. In how many hours can each digger dig a ditch?

2. Solve the equation

$$\sqrt{3 + 2 \text{tg} x - \text{tg}^2 x} = \frac{1 + 3 \text{tg} x}{2}.$$

3. Solve inequality

$$\log_{\log_2} x \left( \log_2^2 x - \frac{7}{2} \log_2 x + \frac{45}{16} \right) < 1.$$

4. Solve inequality

\frac{\sqrt{2}}{2} \cos x - \cos^2 x}{4 \sin^2 x - 3} < 0.

5. Find the volume of a regularly truncated triangular paramid if its side greater than the base is equal to $a$, the dihedral angle formed by the side face and the plane of the base is equal to $\pi/6$, and the height of the paramid is equal to $a/12$.

Option 3

1. Two dump trucks had to transfer the cargo in 3 hours 20 minutes, but the second one was late and arrived at the loading site when the first one had already transferred 2/5 of the total cargo. After that, only the second dump truck carried the remaining cargo. Transporting the entire cargo took 8 hours. In how many hours can each dump truck individually transfer the entire load?

2. Solve the equation

$$\sqrt{\cos(\cos x)} = 1 - 2 \cos(\cos x).$$

3. Solve inequality

$$\log_{\frac{1}{3}} x \left( \log_2^2 x - 19 \log_{\frac{1}{3}} x + 84 \right) < 1.$$

4. Solve inequality

\frac{\sqrt{3}}{2} (\cos x - 1) + \sin^2 x}{4 \sin^2 x - 3} < 0.

5. A ball is described near a regular triangular paramid from the side of the base $a$ and the angle of inclination of the side edge to the plane of the base $\alpha$. Find its volume.

Option 4

1. Two quantities left simultaneously towards each other from points $A$ and $B$ and met after 1.2 hours. How many hours did each of them travel the distance $AB$ if the first arrived at $B$ one hour later than the second at $A$.

2. Solve the equation

$$|\cos 2x| = \left| \sin^2 x - \frac{1}{4} \right|.$$

3. Solve inequality

$$\log_{\frac{1}{2}} \frac{x^2 - 3x + 2}{4x - 10} < 0.$$

4. Solve inequality

$$\frac{1 + \sin \frac{x}{2} - \sqrt{3} \cos \frac{x}{2}}{4 + 4 \sin x + 2 \cos x + \sin 2x} > 0.$$

5. At the base of a quadrangular paramid lies a rhombus with sides $a$ and acute angle $\alpha$. All side faces are inclined to the base plane at an angle $\beta$. Find the complete surface of the paramid.

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How we solve the equation! Based on this equation, we write in a row one, another, third, … - a chain of equations in which each is obtained from the previous transformation of one or both parts at once, and so on until we get a completely simple equation, the solutions of which we already know how to find. If, in this case, all transitions from this equation to the next are “reversible”, that is, we see that the transformation of the fourth equation can be obtained by the third, from the third - the second, etc., then it is clear that the last process of the equation has the same roots as the original one. But we often encounter such equations with radicals, logarithms, and trigonometric functions, the solution of which involves resorting to rather complex transformations, so that the “reverse transition” is not very clear. If, when solving such equations, you make transformations formulaically, without delving into their meaning, then it is easy to make a mistake: lose a root or add an extra one. This article explains how this happens and how to avoid such troubles.

1. SOME NOTATIONS.

Let the data be number sets $X$ and $Y$. The notation $X \subset Y$ means that each element of the set $X$ is at the same time an element of the set $Y$, and in $Y$ there may be elements that do not belong to the set $X$. The notation $x \in X$ (respectively $x \in X$) means that the element $x$ belongs (does not belong) to the set $X$.

The conditions below together with the phrase “the set $X$ consists of all real numbers $x$” also use the notation: $X = \{x| -\infty < x < \infty\}$. Similarly, the notation $X = \{x|x \ge 3\}$ means that the set $X$ consists of all those numbers $x$ that satisfy the inequality $x \ge 3$, and the notation $X = \{x|x \neq \pi\}$ means that the set $X$ contains all real numbers except $x = \pi n^*$).

And, finally, the conditions for the range of permissible values ​​of the VA (recall that the range

*) Throughout what follows it is assumed that $n, k=0, \pm 1, \pm 2, \ldots$

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The admissible values ​​of the equation $f(x)=g(x)$ are the set of those $x$ for which the function $f(x)$ and $g(x)$ are defined) the given equation is denoted by the letter $X$, and the OD of the equation obtained from the original one as a result of some transformation is denoted by the letter $Y$.

2. WHERE DO THIRD-PARTY ROOTS COME FROM?

The first reason is the expansion of ODZ. When solving an equation, a variety of transformations are constantly applied. Quite often, it happens that the OD of the equation obtained from the original one as a result of some transformation turns out to be wider than the OD of the original equation (the phrase “wider OD” means $X \subset Y$). Expansion of the ODZ can lead to the appearance of permanent roots. The point is that the resulting equation may have roots that, being elements of the set $Y$, do not belong to the set $X$. They will be constant for the original equation.

Example 1: Solve the equation
$$\frac{2}{2-x} + \frac{1}{2} = \frac{4}{2x-x^2}.$$
Solution. Let us first write down the ODZ of equation (1): $X=\{x|x \neq 0;2\}$. Reducing to a common denominator, we get
$$4x+2x-x^2=8.$$

Equation (2) already has another ODZ: $Y=\{x|-\infty < x < \infty\}$. Solving equation (2), we obtain two roots: $x_1=4$, $x_2=2$, with $x_1 \in X$, while $x_2 \in X$, and therefore $x_2$ is a constant root for equation (1). Thus, equation (1) has a single root $x=4$.

Example 2: Solve the equation
$$\lg (x^2+3x-4) = \lg (2x+2).$$
Solution. Having solved the system of inequalities
$$\begin{cases} x^2+3x-4 > 0, \\ 2x+2 > 0, \end{cases}$$
we find the ODZ of equation (3): $X=\{x|x > 1\}$. By potentiating, we transform equation (3) into the equation
$$x^2+3x-4=2x+2$$
with wider ODZ: $Y=\{x|-\infty < x < \infty\}$. Equation (4) has two roots: $x_1=2$, $x_2=-3$, with $x_1 \in X$ and $x_2 \in X$. This means that $x_2$ is a constant root for equation (3) and, therefore, equation (3) has a single root $x=2$.

The expansion of ODZ sometimes results from the use of certain formulas known from the secondary school course. These are formulas related to various properties of logarithms, radicals of even degree, as well as certain

$$\log_a f(x) + \log_a g(x) = \log_a [f(x) \cdot g(x)],$$
$$\frac{1-\cos x}{\sin x} = \tg \frac{x}{2}.$$

It is easy to see that a function containing these equalities on the right-hand sides has a wider domain of definition than a function containing on the left-hand sides.

Thus, the domain of definition of the function $\log_a f(x)+\log_a g(x)$ is determined by the system of inequalities
$$\begin{cases} f(x) > 0, \\ g(x) > 0, \end{cases}$$
whereas the domain of definition of the function $\log_a [f(x) \cdot g(x)]$ consists of solutions to system (7) and solutions to another system of inequalities
$$\begin{cases} f(x) < 0, \\ g(x) < 0. \end{cases}$$

The domain of definition of the function $\frac{1-\cos x}{\sin x}$ is: $X=\{x|x \neq \pi n\}$. The function $tg \frac{x}{2}$ has a wider scope of definition: $X=\{x|x \neq \pi(2k+1)\}$. Therefore, when solving an equation, using formulas (5), (6) and the like “from left to right” can lead to the appearance of constant roots.

Example 3: Solve the equation
$$\frac{1-\cos x}{\sin x} + \tg^3 \frac{x}{2} = 0.$$

Solution. Let us write down the ODZ of equation (8): $X=\{x|x \neq \pi n\}$. Applying formula (6), we obtain the equation
$$tg \frac{x}{2} + \tg^3 \frac{x}{2} = 0$$

with wider ODZ: $Y=\{x|x \neq \pi(2k+1)\}$. The solutions to equation (9) are the values ​​of $x=2\pi n$. All these values ​​are not included in the ODZ of equation (8), and therefore are constant roots. Thus, equation (8) has no solutions.

The second reason is the reduction of both sides of the equation to the same function. Let me give the equation
$$f(x)=g(x).$$
Let’s multiply both sides by the function $h(x)$. Let’s learn the equation
$$f(x)\cdot h(x)=g(x)\cdot h(x),$$
the roots of which are both the roots of equation (10) and the roots of the equation $h(x)=0$ (or, as they also say, the roots of the function $h(x)$. The averages of these latter may also be roots that do not satisfy the equation

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(10), that is, it is a construction for it (by the way, in example 1, both sides of equation (1) were multiplied by the same function $2x(2-x)$. As a result, the constructed root $x=2$ appeared, but we owe its appearance to expanding the ODZ).

Example 4: Solve the equation

$$x = (\sqrt{1+x} + 1)(\sqrt{10+x} - 4).$$

Solution. Let us write down the ODZ of equation (12): $X = \{x|x \geq -1\}$. Let’s multiply both sides of the equation by the function $h(x) = \sqrt{1+x} - 1$. After the transformation, we obtain the equation $x(\sqrt{1+x} - \sqrt{10+x} + 3) = 0$, which has 2 roots: $x_1 = 0$, $x_2 = -1$ (note that the extended ODZ will not occur). Both roots belong to the set $X$, however, verification convinces us that $x_1 = 0$ is a constructed root (it is easy to see that $x = 0$ is the root of the function $h(x)$). Thus, equation (12) has a single root $x = -1$.

The third reason is the construction of both sides of the equation in the central one. Consider equation (10). Let’s square both sides of this equation. We obtain the equation $[f(x)]^2 = [g(x)]^2$, the roots of which are both the roots of equation (10) and the roots of the “extraneous” equation $f(x) = -g(x)$. The averages of these latter may also turn out to be roots that do not satisfy the original equation (10). They will be constructed for equation (10).

Example 5: Solve the equation

$$\sqrt{2x+5} + \sqrt{x-1} = 8.$$

Solution. Let us write the ODZ of equation (13): $X = \{x|x \geq 1\}$. Let’s square both sides of the equation. After transformation we get

$$2\sqrt{2x^2 + 3x} - 5 = 60 - 3x$$

and onwards

$$4(2x^2 + 3x - 5) = (60 - 3x)^2,$$

whence $x_1 = 10$, $x_2 = 362$. Consequently, the solutions found, we replace that only $x_1 = 10$ satisfies equation (13), while $x_2 = 362$ is the constructed root (although it belongs to the ODZ of equation (10). The extraneous root appeared as a result of squaring both sides of equation (14).

3. WHAT CAUSES ROOTS LOSS?

The first reason is the narrowing of the ODZ. There are also transformations that lead to an equation with a narrower OD. Narrowing the ODZ can lead to the loss of roots, namely, those roots of the original equation that belonged to the set $X$, but do not belong to the “narrower” set $Y$, $(Y \subset X)$, can “lose”.

Example 6: Solve the equation

$$\text{tg} \left( x + \frac{\pi}{4} \right) = 2 \text{ctg} x - 1.$$

Solution. Let us transform equation (12) to the form

$$\frac{\text{tg} x + \text{tg} \frac{\pi}{4}}{1 - \text{tg} x \cdot \text{tg} \frac{\pi}{4}} = \frac{2}{\text{tg} x} - 1.$$

Putting $y = \text{tg} x$, we arrive at an algebraic equation for $y$, solving which we obtain $y = \frac{1}{2}$, that is, $x = \text{arctg} \frac{1}{2} + \pi k$.

But verification convinces us that the values ​​of $x = \frac{\pi}{2} + \pi n$ also satisfy equation (15). When did we lose these roots? Let us compare the ODZ of equations (14) and (15). X consists of all values ​​of $x$ except those for which $ctg x$ and $\text{tg} \left( x + \frac{\pi}{4} \right)$ do not exist. Means,

$$X = \left\{ x \mid x \neq \pi n, x \neq \frac{\pi}{4} + \pi n \right\}.$$

The ODZ of equation (16) - $Y$ - consists of all values ​​of $x$, except for those for which $ctg x = 0$, $\text{tg} x = 1$ and $\text{tg} x$ do not exist. Means,

$$Y = \left\{ x \mid x \neq \pi n, x \neq \frac{\pi}{4} + \pi n \right\}.$$

It follows that $Y \subset X$, that is, the ODZ of equation (16) is narrower than the ODZ of equation (15). The values ​​$x = \frac{\pi}{2} + \pi n$ belong to the set $X$, but do not belong to the set of roots of the original equation.

Now, one important question arises: what led to the narrowing of DL? Analysis of the solution to Example 6 shows that the use of tangent sum formulas led to narrowing the OD; in our example, the formula belonged to the set $X$, but did not belong to the “narrower” set $Y$, $(Y \subset X)$.

The second reason is dividing both sides of the equation by the same function. Let $x_1$ be the root of equation (11), but it does not necessarily satisfy equation (10). Thus,