Kvant 1970, No. 7 · Pages

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produces larger or smaller sizes. Thus, “houses” really exist, but “material points” do not.

The same can be said about the geometric concept of “segment”. In nature, we do not observe “in their pure form” objects that are long and “straight,” but “devoid of thickness.” This concept, therefore, does not arise as a result of the abstraction of identification, but is an idealization.

Idealization in general terms is as follows. The real object is replaced in our minds by an abstract model, which is obtained not only by abstracting from non-essential properties, but also by endowing the model with idealized properties, which in an exaggerated form reflect the essential properties of the object. For example, if we study the movement of a celestial body distant from us at a distance many times greater than its size, we can build an exaggerated model, considering this body to be a material point. In the same way, the concept of “touch” in an exaggerated form reflects the possibility in reality of the relative position of objects.

Of course, this does not mean that the concept that arose as a result of idealization is completely divorced from the objective world and devoid of real meaning. Allowing us to especially deeply and clearly trace the most essential properties of an object, these concepts have a well-known significance in practical applications of mathematics.

A special type of abstraction, idealization, is the abstraction of local feasibility. It is characterized by a distraction from the real boundaries of our constructive capabilities, conditioned by our limited life in space and time.

This form of abstraction allows us to reason, for example, about any large number (say, about the number $10^{10}$) and assume that we seem to have the opportunity to reach it, starting with 1, 2, 3, 4 … etc. Obviously, this possibility is purely potential: we will not have enough for this throughout our entire life.

However, we abstract from the practical impracticability of this task, from the limitations of our capabilities, and consider it permissible to consider any numbers, thereby arriving at the concept of “an infinite series of natural numbers.”

As a result of the use of this type of abstraction, concepts such as “straight” (extending unlimitedly in both directions), “plane” and others also arise.

Modern mathematics operates with a huge number of concepts that cannot be obtained as very strong, immediate abstractions of the properties of an object in the mathematical world accessible to animal contemplation. These concepts are formed in the process of multi-stage layering of abstraction.

The original, primary mathematical concepts arise from objective reality, reflecting the abstraction of quantitative relations and spatial forms of the object of action. But when these concepts arise, they seem to live their own life, forming their own world in the human mind, and they themselves give rise to further reflections and new abstract concepts. These abstract concepts, in turn, again give rise to new abstractions, so to speak, of a higher order, etc.

In order to understand what has been said, let us turn to the concept of “numbers”. Number, which initially acts as a result of counting objects (identification abstraction), is then generalized to the concept of a natural number (potential feasibility abstraction). Then the concept of a fractional number (identification abstraction) is introduced.

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Fig. 15.

development of a cylinder with infinite height and would require an infinite sheet of paper for its image. Therefore, the circumpolar regions are not depicted on such maps (the upper and lower boundaries of the map, of course, are determined purely arbitrarily, from messages of convenience; for example, figure 15 is limited from above by the 85th latitude, and by the 30th latitude).

And two more comments. Firstly, our reconstruction of Mercator’s train of thought may not at all coincide with the actual course of his reasoning; It just seemed to us that this was the easiest way to talk about it. Further, it should be noted that the conclusion about the stretching of the map along the meridians by $\frac{1}{\cos \varphi}$ times, which we obtained “on our fingers”, for a strict (although not complicated) justification would require knowledge of the elements of mathematical analysis, including natural logarithms and the beginnings of differential and integral calculus *). By the way, Mercator himself actually did not do this - a rigorous solution to the problem of “straightening rhoxodromes” was obtained only thirty years later by Richard

*) This information can, for example, be found in the books by P. Courantan G. Robbins “What is mathematics?” (M., 1967) or N. Ya. Vilenkina S. I. Shvartsburg and “Mathematical Analysis” (M., 1969).

Wright from Cambridge (whose editors publish modern editions of what we call the “Mercator projection”).

Lambert cylindrical projection

Let’s consider another projection with a rectangular grid; it turns out very simple. Let us describe a right circular cylinder around the globe (contacting it along the equator and in stripes) and project onto its lateral surface the surface of the globe (without stripes!) so that the projection ray for each point of this surface is the passage through it of a perpendicular to the earth’s axis (Fig. 16).

Using the usual formulas of school stereometry, it is easy to show (try it!) that the areas of any sections of the surface of the globe and their projections onto the lateral surface of the cylinder turn out to be equal, and therefore the development of this cylinder (cut along any meridian) is called an equal-volume cylindrical projection (or Lambert’s projection). Correctly conveying areas, the Lambert projection greatly distorts the shape of even small sections of the projected surface (compare the ellipses on figure 17 with the corresponding circular ones on figure 15): the circumpolar regions are solidified here

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Fig. 17.
stronger than in the square projection, from which the Lambert projection can be very much obtained by compression along the vertical axis. The straightness of the images of meridians and parallels of the perfect is obvious.

Orthodromi

Until now, we have taken it almost for granted that the simplicity of plotting a course on a map is important to sailors, but the distance that a ship has to travel is of much less interest to them. Well, during the time of sailing ships, sorry, that’s how things were. But nowadays, when sailing over very long distances, not lasting at all for months, and for years, not to mention long-distance (including transcontinental and international) aviation flights, the issue of minimality of the trajectory becomes of paramount importance.

It can be proven (which we will not try to do here) that on the surface of a ball the shortest line connecting two arbitrary points of it (or, as is commonly said in mathematics, a geodesic) is the lesser of each of the great circle passing through these points. (For diametrically opposite points on the sphere, both arcs of the great circle passing through them are equal, and

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There may be a discrepancy between the lokompnya and the orthodrome. Of course, with long flights the difference may be greater when traveling in temperate latitudes; for example, the path along the locodrome from Moscow to New York is 8500 km, and along the orthodrome - 7500 km.

Gnomonic projection

A projection in which orthodromes are depicted as straight lines is obtained by central projection from the center of the earth’s sphere: in this case, all rays directed to the points of the orthodrome (which, as we saw, are a large circle) lie in the same plane, intersecting in a straight line with any projection plane. The grid of meridians and parallels looks easiest in this so-called gnomonic projection (Fig. 19), if the projection plane touches the globe projected onto it at the pole. (A gnomon is the rod of a sundial; if we insert it onto the projection plane and consider the earth’s radius to be a gnomon, then the projection of the projecting ray will indeed appear to us as the shadow of this gnomon.)

It is easy to prove (do it yourself) that meridians are depicted in a gnomonic projection of this type as straight lines, and parallels as concentric circles. On figure 19 (where, for obvious reasons, there is no image of the screen: it “gap into infinity”) the spiral-shaped form of locodromy in the vicinity of the pole in the gnomonic projection is very clear.

For sea and air travel over long distances

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Gnomonic projections are very convenient. Of course, the entire earth’s surface cannot be depicted on such a map, so one has to choose different projection planes, the position of which determines the shape of the coordinate grid depicted on them. Figures 20, a and 20, b depict gnomonic projections, the picture plane of which touches the globe not at the pole.

Task. Let the terrain plan be depicted by projecting a given section, shaped like a “cap” with a radius of 100 km, onto a plane tangent to the “cap” in the center ($R_{\text{земли}} \approx 6000$ km). Find the accuracy of the image, i.e. find out by what percentage small segments of meridians and parallels drawn at any point of the “cap” are distorted:

a) in gnomonic projection (central projection from the center of the Earth);
b) in orthographic projection (parallel projection in a direction perpendicular to a given tangent plane). See figure 21, a, b.

Instead of a conclusion
That’s the whole story for now about the basic principles and specific ways of depicting the earth’s surface on a map. For anyone who would like to get acquainted with the science of “portraits of the Earth” - cartographies - more deeply, in more detail and systematically, we would advise, first of all, to carefully review some large atlas, paying special attention to the difference in the projections used in it, and try to understand independently (with the help, perhaps, of rather brief notes usually given in atlases) the reasons for this difference. True, in doing so, you may be enriched not so much with answers, but with new questions. But so much the better!

You will find a systematic (but very rich and therefore much more difficult than in this country) coverage of the subject in the article “Cartographic Projections” in the latest edition of the Great Soviet Encyclopedia. In the article “Geographical Maps,” very rich in historical material, placed there, you will find further references to the literature.

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Metals are opaque, shiny and comparatively hard. Metals are strong, but their shape can be changed through forging and rolling. They can also be floated and welded. Metals have good electron and thermal conductivity. They owe all this to the metal connection. The nature of this connection is that in a metal each atom is surrounded by a large number of points of the same atoms, each of which has only a few electrons in its outer electron shell, and the electron shells overlap so much that the weakly held outer electrons can no longer be attributed to any one of the atoms. While the atoms, or rather ions, remain in place, the “electron gas” moves freely among the ions, bonding them to each other like glue.

Due to the fact that electrons are free and can mix in an electric field, metals are conductors. Because free electrons absorb and then re-emit most of the light that hits them, metals are opaque and shiny. Due to the fact that free electrons can transfer thermal energy, metals have high thermal conductivity.

This article does not discuss the thermal, electrical, or optical properties of metals. I will mainly talk about their mechanical properties.

The nature of the metallic bond is the same for different metals. It promotes the formation of dense packing of atoms, characteristic of

In the image (see Screensaver), obtained with the help of a different microscope, the thickness of the boundaries of the constellations is only a few atoms. The photograph shows the tip of a tungsten needle, magnified approximately 5 million times. Each bright dot corresponds to a tungsten atom. The location of the dots depends on how the game surface intersects the crystal plane. A violation of the structure at the grain boundary is visible, going from the lower left to the upper right corner of the image.

Such structures, while resisting rolling, provide less resistance to the slug, which is why the metals are plate-like. Dense packaging also requires a high specific gravity of metals. The mechanical properties of metals are therefore determined by the crystal structure that is most suitable for metallic bonding carried out by free electrons.

Back in 1665, Robert Hooke modeled the characteristic shape of crystals using blocks laid in rows, but only 250 years later it was discovered that he had built an accurate model of the crystal structure of known metals, in which each atom corresponds to thermal energy; metals have high thermal conductivity.

Although it has been known for several centuries that some substances have a crystalline structure, the fact that ordinary metals are crystals remained in doubt until recently. Properties such as the solidification of the melt at a very specific temperature and the presence of shiny grains on the surface of fractures spoke of a crystalline structure, but other facts seemed to indicate amorphism: molten metals after solidification took the shape of a vessel, solid metals were plate-shaped and their shape could be changed by forging, the polished surface of the metal looked completely uniform.

The key to elucidating the structure of metals was found in 1864, when the Englishman Henry Sorby developed a method for studying metals under a microscope using reflected rather than transmitted light. In addition, he owed his success to the fact that, in addition to the polished surface, he examined the metal surface treated with weak chemicals.

*) On the dense packing of atoms, see the article by G. I. Kosurov “Crystals from balls”, «Kvant» No. 1.

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mic reagent. Some reagents etched deep grooves along the grain boundaries, others revealed the microstructure of the grains, acting on the grains themselves.

This is an opportunity to detect an irregular network of boundaries that split the metal into small polyhedra - grains about 0.25 mm in size (Fig. 1).

Under the same lighting, some grains appeared brighter, others darker. The distribution of light and shadow varied with varying angles of illumination, showing that the etched surface of each grain was composed of small, flat, reflective steps unequally inclined toward the metal surface. It became clear that each grain, despite its irregular shape, is a separate crystal, and the entire piece of metal consists of many such small crystals, oriented in different directions and having common boundaries.

The metallography method invented by Sorbi turned out to be very fruitful for determining the shape and size of grains, which is important for studying the technical properties of metals, for detecting impurities and their accumulations along grain boundaries, for studying alloys. However, optical metallography could not give

Fig. 1.

Micrograph of zeren aluminum in illuminated light. Etching sample. As shown schematically on the right, some grains are oriented so that they appear bright, others - dark. Magnified 60 times.

information about the structure of grain boundaries. Gradually, researchers came to the idea that when a metal cools below its melting point, neighboring grains capture as many atoms as possible, reducing the boundary between grains to a layer only a few atoms thick, within which the crystallographic orientation changes sharply from one grain to the next.

Recently, this point of view was confirmed by the use of an inon microsop, invented by the American Erwin Muller. In this device, the tip of a metal needle is held at a high positive potential in a vacuum, so that the electric field lines start at the needle and end at a fluorescent screen. A small amount of helium is allowed into the chamber. When a helium atom collides with a needle atom, it becomes positive and flies towards the screen along the drain line. When removed from the screen, a visible image is formed. The size ratio is such that the area of ​​one atom at the tip of the needle corresponds to approximately one square millimeter on the screen; It is in this sense that we “see” the atomic structure of the needle (see photo at the beginning of the article).

In 1900, James Ewing and Walter Rosenhein discovered that

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If the sample is subjected to deformation, for example, by shifting its faces in different directions, then its surface is covered with lines. They usually run strictly parallel to each other within each grain, and their direction is different in different grains (Fig. 2). Careful observations showed that these lines are traces of steps formed when thin layers of crystal slide over each other. Further studies (especially with large ideal crystals of metals) showed that this sliding occurs along certain planes and in certain crystallographic directions. Interim plastic deformation of metals thus turns out to be very different from flowing liquids and gases. Plastic deformation is characteristic specifically of the crystalline structure. It happens in such a way that atoms on one side of the slip plane are separated from their previous neighbors on the other side of the plane and move, carrying with them a certain part of the crystal and finally occupying a position in which they, together with their new neighbors, form a columnarly regular structure, as before. In this case, the original properties and internal structure of the crystal are restored.

If plastic shear is a trace

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sewing one dense, close-packed flat structure over another. Each three adjacent balls in the plane form a hole in which the ball of the top layer can be placed.

There are two ways to select such holes. If the locations of the balls of the first layer are denoted by the letter A, and the two subsequent layers by the letters B and C, respectively, then a face-centered cubic lattice is obtained if the layer is placed in the sequence ABCABC, and a hexagonal close packing - with ABABAB.

In metal crystals, sliding occurs along those directions where the atoms are packed most densely, since the resistance to the movement of the layers relative to each other in this case is less. In addition, as we will see later, atoms moving along close-packed rows more often occupy the source positions. Because they are more symmetrical, cubic structures contain many such directions in which rows of atoms are tightly packed and planes of atoms can slide against each other in many directions. This greatly affects the flatness. Individual crystals can take such a shape that neighboring grains fit close to each other, without leaving gaps or cracks, and, therefore, the entire crystal can also take any shape without breaking. Crystals with a hexagonal structure are more fragile and less amenable to mechanical processing.

The plastic properties of crystals are usually determined, as we have seen, by the origin of the close-packed directions and planes along which the fusion occurs. Why do such crystalline structures form? To answer this question, we must turn to the electronic structure of metals.

In 1900, the German physicist Pavel Drude proposed that a metal contains free electrons that can move throughout the entire volume of the metal under the influence of an electric field, which is the reason for the high electrical conductivity of metals. This theory was subsequently improved by the application of quantum mechanics, but the basis remained the same: a gas of mobile electrons, like liquid glue, holds the positive ions of the metal together due to the forces of electrical attraction to this gas. Electron gas ions bend towards each other, trying to form a compact mass, the structure and volume of which is determined mainly by the geometry of close-packed balls. If the balls are identical, as in pure metal, then simple crystalline structures arise,

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discussed by us earlier. In some alloys, different structures can be formed, and even denser ones, due to the difference in the size of the balls.

Because the electron gas behaves like a glue that binds nonions, the valence of metals does not affect their crystal structure, as is the case with nonmetals. Crystals of metals can be so strongly interconnected by a gas of free electrons that the boundaries between them are almost invisible. It is very difficult to break cold metal along grain boundaries unless the boundaries are weakened by impurities. This “promiscuity” of the metal bond leads to the fact that two pieces of metal can be connected to each other by simply pressing their clean surfaces. metallic bonding makes it possible to obtain alloys of various metals in very diverse combinations and proportions.

How does the presence of free electrons affect the strength and ductility of metals? We have already considered one effect: the formation in metals (and in some alloys, in which the sizes of the atoms do not differ too much from each other) of simple crystalline structures with densely filled planes and lines along which sliding occurs. The plasticity of metals is also explained by the fact that atoms are not directly connected to each other, but are attracted to one another due to the gas of free electrons. Therefore, sliding of one layer relative to another will occur easily.

If the sliding of the layers occurred in the complete absence of resistance, then the material would not have any rigidity at all. The solid body, however, has a well-defined stiffness, the measure of which is the shear pattern, which determines the reinforcement required for very small deformation.

Consider two fusion planes (Fig. 4). In one raft (a) the atoms in each direction are more densely packed, and, therefore, the distances between planes in the vertical direction are smaller than in raft b. It is clear that the same deformation in raft a will require less effort than in raft b. For clarity, we can say that it will take less xmlns to “climb” (in reality, this means that less force will be required to deform the electron shells). In other words, along densely packed planes the shear pattern is smaller.

The strength of a metal is also characterized by the amount of deformation that causes its flow -

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Fig. 5.
A crystal with a regular lattice can be thought of as a package of parallel atomic planes (a). If one or more planes rotate inside the crystal, then the edges of the “extra” planes form an edge dislocation in the crystal (b). Another simplest type of defect is a screw dislocation. Here, none of the atomic planes are sliding of atoms. In the case of sliding planes, plastic deformation begins when the atom of the upper layer approaches the position of unstable equilibrium, that is, when the difference between the electrical forces tending to return the atom back and the electrical forces pushing it forward to the next “hole” is maximum. It is clear that in case a this deformation is smaller. Because of these two reasons: less resistance to shear and less deformation required to reach the yield point, sliding is more likely along closely packed planes. These reports explain why metals with a cubic structure, such as copper and aluminum, are particularly ductile, but the question of strength remains open. Calculations based on these reports indicate that metals must be elastically deformed by 3-10% before yielding occurs. However, pure metals already have a deformation of 0.01%.

A discrepancy of a thousand times is of both scientific and technical interest. When this discrepancy was first discovered, the entire theory initially seemed to be wrong. But this is not entirely true, or metal crystals grown in the form of thin “whiskers” have a strength close to the calculated value. The bones do not roll inside the crystal, but the planes themselves near the linear dislocations are no longer parallel, but are closed with each other so that all crystals seem to consist of a single internal curved atomic plane (b). When bypassing a line dislocation, this plane rises (or falls) by one pitch equal to the interplanar distance.

The lower strength of large crystals is explained by the onset of dislocations in the crystals - violations of the regular structure, which give the atomic planes the opportunity to slide relative to each other more easily than in an ideal crystal.

Suppose we need to knit together a larger, heavy carpet. If you try to pull the entire carpet together, the resistance will be too great. But you can make a fold in the carpet and move it. This action does not require a lot of energy, and as a result, the weight of the carpet is pulled together.

The movement of the carpet in this case is very similar to plastic shear caused by dislocation. However, the crystal will slide over the other not entirely, but in parts. During such a sliding, a boundary line (dislocation) will necessarily appear, separating the part that has already fused from the part that has not yet fused.

A distinction is made between “edge” and “screw” dislocations (Fig. 5). In the first case, the line dislocation is perpendicular to the sliding direction, in the second case it is parallel. Most dislocations are a combination of these two types and are shaped like a spring.

Dislocations in a crystal are almost inevitable, if only because of the unevenness of the crystal growth process and the presence of grains, the boundaries of which

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Fig. 6.
The dislocation moves along the crystal. The diagram or will construct the process of plastic svir in a close-packed crystal. Under the action of the twisting force (arrow), the atomic plane (black balls) mixes (2) and binds (dotted line).

The extra plane (black balls) continues to move, the connections are broken and then restored (black line). When a dislocation (area of ​​broken bonds) passes through the entire crystal, it becomes deformed (6).

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Fig. 7.

Point defects, like dislocations, disrupt the structure of the crystal. There are three types of precise defects (from left to right): vacancy (absence of an atom), substitution (inclusion of an atom of a foreign substance).

The result is actually a collection of dislocations. Shear forces applied to the crystal cause the dislocation to move along the slip planes. If there is only one dislocation in the crystal, then during shear it leaves the crystal (Fig. 6). In reality, a crystal usually contains a complex network of dislocations connected to each other, not counting other structural disturbances and the presence of impurities. Since the ends of the dislocation are fixed either to other dislocations or to impurities, when sheared the crystal does not get rid of disturbances in its structure. In fact, the number of dislocations increases with shear.

Dislocations move along the slip plane due to “shear forces” that occur during stretching, rolling, or torsion of the sample. What is the amount of force required to move a dislocation? This question contains two questions: 1) about the natural resistance to the movement of a dislocation in an ideal crystal and 2) the influence of interference (impurities and other obstacles) found in a real crystal.

Let us consider the resistance to dislocation motion in an ideal lattice. The atoms located immediately in front of the dislocation repel it, since it tends to become interstitial atom (extra atom). Each of these defects disrupts the crystal lattice in its own way (red lines). Fine defects often interact with dislocations.

remove them from a position of stable equilibrium. Atoms located behind the dislocation push it forward, as they strive to occupy a new stable position. The dislocation is subject to equal and opposite forces, so the resistance to its movement through the crystal is zero! This strange property of the crystalline state occurs if the dislocation area is large enough. In this case, there are so many atoms on both sides of the dislocation, but their action completely balances each other. Otherwise, force must be applied to move the dislocation. In the limit where the thickness of the dislocation does not exceed the size of an atom, this force is almost equal to the readability of the materials.

Narrow dislocations can occur in crystals such as diamond, so such materials even with dislocations are readable by scales. Wide dislocations explain the softness of metals such as copper, gold, and aluminum. The practical problem with these metals is not to make them softer, but rather to make them hard. Metallurgists achieve this by placing various obstacles in the path of dislocation movement (Fig. 9).

The introduction of impurity atoms leads to local disturbances

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Fig. 8.
Single dislocation in a cadmium crystal. The simok was obtained using an electron microcoppe and a special photography method.

crystal structures. These disturbances resist the movement of dislocations. The effect of impurity atoms is enhanced when they are grouped. This can be achieved by heat treatment. Since dislocation is concentrated at grain boundaries, the metal can be simplified by reducing the grain size.

If there are many dislocations, then when they move along the slip planes they interfere with each other—this effect can be easily imagined by anyone who has had to be stopped at a street intersection with heavy traffic.

Let’s look at the last question. What happens when metal tries to cut quickly? Obviously metals contain impurities of various brittle substances, and if the frequency of such a substance is revealed, then the crack quickly extends beyond the inclusion. Experience shows that metals with a face-centered lattice, such as copper, resist cracking better. It does not spread throughout the corn, but “fades out” due to plastic deformation.

Fig. 9.
Impurities can pin a dislocation, causing it to bend and cross different planes, which interferes with the movement of the dislocation. The micrograph obtained using an electron microcoppe shows dislocations attached to impurities in a cadmium crystal. Magnified 90,000 times.

Metals with a body-centered lattice, such as iron, behave like copper when heated, but open easily when cold.

If the wedge moves slowly, then the dislocations already present in the metal move under the influence of the direction that appears in the air when the wedge moves, and its energy is wasted on plastic deformation. What happens if the wedge moves quickly?

Kelly, Tyson and I recently addressed this issue. We calculated that immediately near the sharp edge of the wedge, the force tending to break atomic bonds is 5-6 times greater than the force tending to cause shear. Both forces increase as the wedge approaches, but their ratio remains the same until the limit of the strength of atomic bonds to break or shear is reached.

If the ratio of the tensile strength to the shear strength is greater than the ratio of the tearing forces to the forces causing the displacement of atoms, then the failure will have the form of sliding deformation. If it’s the other way around, the material will crack.

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Fig. 10.
Three dislocations modeled using soap bubbles. (A layer of bubbles models a layer of atoms.) Dislocations, each of which contains an extra row, run diagonally from the lower left to the upper right corner.

Fig. 11.
A wedge moving through a metal at high speed can either simply break the bonds between atoms or cause the atoms to slide over each other. In brittle material (above), the bonds break early, the crack spreads quickly, and the metal falls apart. In the vz material (below), when the wedge moves, shear will occur. Bonds break and are restored after atomic displacement. Klin respects.

Tensile and shear bond strengths can be assessed for various materials. If this is done, it turns out that materials like diamond must be brittle. Metals with a volumetric cubic lattice can be either brittle or brittle. In metals with a grated lattice, the shear bond strength is called

Abridged translation by E. M. Dijur (A. H. Cottrell, The Nature of Metals. “Scientific American”, September 1967, p. 90).

A. H. Cottral

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The Case of the Method of Mathematical Induction

One day, a teacher gave 10th grade students the following homework problem:

To prove, by attempting to use the method of mathematical induction, that for any natural number $n$ the inequality is true

$$P_n = \frac{1 \cdot 3 \cdot 5 \dots (2n-1)}{2 \cdot 4 \cdot 6 \dots 2n} < \frac{1}{\sqrt{n+1}}.$$

Kolya V. found the following solution at home;

a) For $n = 1$ the inequality is true, since

$$P_1 = \frac{1}{2} < \frac{1}{\sqrt{2}}.$$

6. Let us show that the validity of inequality $P_{n+1} < \frac{1}{\sqrt{n+2}}$ follows from condition $P_n < \frac{1}{\sqrt{n+1}}$. To do this it is enough to show that

$$\frac{P_{n+1}}{P_n} < \frac{\sqrt{n+1}}{\sqrt{n+2}},$$

or

$$\frac{2n+1}{2n+2} < \sqrt{\frac{n+1}{n+2}},$$

or

$$(2n+1)^2(n+2) < 4(n+1)^3.$$

or

$$4n^3 + 12n^2 + 9n + 2 < 4n^3 + 12n^2 + 12n + 4,$$

or

$$0 < 3n + 2,$$

which is obviously true. Thus, a fairly provable inequality follows from the principle of mathematical induction.

However, while telling this solution at school, Kolya forgot to put one in the denominator and began to prove a more “gross” inequality:

$$P_n < \frac{1}{\sqrt{n}}.$$

Here’s what he got:

a) For $n = 1$ the inequality is true, since

$$P_1 = \frac{1}{2} < \frac{1}{1}.$$

6. Let us show that condition $P_n < \frac{1}{\sqrt{n}}$ implies $P_{n+1} < \frac{1}{\sqrt{n+1}}$. To do this it is enough to show that

$$\frac{P_{n+1}}{P_n} < \frac{\sqrt{n}}{\sqrt{n+1}},$$

or

$$\frac{2n+1}{2n+2} < \sqrt{\frac{n}{n+1}},$$

or

$$(2n+1)^2(n+2) < 4(n+1)^3n,$$

or

$$4n^3 + 12n^2 + 9n + 2 < 4n^3 + 12n^2 + 12n + 4,$$

or

$$0 < 3n + 2,$$

which is obviously true. Thus, a fairly provable inequality follows from the principle of mathematical induction.

For those who understand this issue, we propose to prove two inequalities:

1. $P_n < \frac{1}{\sqrt{3,1n}}$.

It is interesting that in this inequality the coefficient 3.1 cannot be replaced by 3.2. The largest coefficient at which it remains true is π (the ratio of the circumference to the diameter).

2. $\frac{1}{2^3} + \frac{1}{3^3} + \dots + \frac{1}{n^3} < \frac{1}{4}$.

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A few words about the author of the solution to Hilbert’s ten problems

Yuri Vladimirovich Matiyasevich was born on March 2, 1947 in Leningrad. The first person who paid attention to the boy’s outstanding mathematical abilities and encouraged him to develop an early interest in mathematics was Sofya Grigorievna Generson, a mathematics teacher at School 255 in Leningrad. Starting from class, Yuri took part in city mathematical optimids, invariably taking first place in them, and in 1961 - 1964 - in the All-Russian and Moscow optimids, where he was also among the winners. (Yuri Matiyasevich graduated from 10th grade at boarding school No. 18 at Moscow State University). From 7th to 9th grade, Yuri studied in the mathematical circle of the Leningrad Palace of Pioneers under the guidance of LOMI employee (Leningrad Branch of the Steklov Mathematical Institute of the USSR Academy of Sciences) N.M. Mitrofanov.

Participating as part of the USSR team in the VI International Mathematical Optimide (Moscow, 1964), Matiyasevich was awarded a 1st degree Diploma.

In 1964-1969. Yuri Matiyasevich is a student of the Faculty of Mathematics and Mechanics of Leningrad University.

While still in his second year, Matiyasevich completed two works on mathematical logic, which were then published in the “Reports of the USSR Academy of Sciences.” The main results of these works were supplemented by Yu. Matiyasevich at the International Mathematical Congress in Moscow in 1966.

Currently, Matiyasevich is a graduate student at the Leningrad Department of the Mathematical Institute. The mathematical creativity of Yu. Matiyasvich is a striking example of the continuity of scientific tradition. Matiyasevich’s supervisor at the university and graduate school is a young but already well-known specialist in mathematical logic, Sergei Yuryevich Maslov, on whose advice Matiyasevich turned to Hilbert’s tenth problem in December 1965. Teacher S. Yu. Maslova prof. Nikolai Aleksandrovich Shanin is the closest student of Corresponding Member of the USSR Academy of Sciences Andrei Andreevich Markov.

Thus, Yuri Matiyasevich is a representative of the “fourth generation”, headed by the outstanding Soviet mathematician Markov, of the so-called constructive direction in logic and mathematics.

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The third issue of Kwan reported on the outstanding success of the young Leningrad mathematician Yu. V. Matiyasevich, who recently managed to take the final step in solving one of the meanings of the “Hilbert problems” posed back in 1900. The work of Yu. V. Matiyasevich was published in the Reports of the USSR Academy of Sciences *). And what happens rarely is that a thoughtful student can independently understand the main content of this work, which is important for modern mathematical science. The fact is that the original part of Yu. V. Matiyasevich’s work is devoted to the proof of the theorem, the formulation of which is quite elementary and is given in the article we publish. The methods of proof are also elementary. It is more difficult to explain the rules in a language understandable to a schoolchild, why this theorem about Fibonacci numbers is of such significant interest for server-side mathematical science. In its formulation and methods of proof, it rather resembles a clever optimization problem.

1. Fibonacci number theorem

Throughout what follows we will deal only with integers. Consider the following property of a pair of numbers (a, b):

“b is divisible by a.”

This property of a pair of numbers can be expressed as follows:

“there is a number x such that ax—b=0.”

We expressed our property of the pair of numbers (a, b) through the existence of a solution to the equation

ax—b=0

(let us recall once again that we are dealing only with integer numbers and, therefore, only with integer solutions). The equation for this is

P(a, b, x) = 0,

where P is a polynomial in three variables a, b, x.

Let us now introduce a general definition; a property of a finite sequence of numbers (a₁, a₂, …, aₘ) is called Diophantine if there exists such a polynomial in m + n variables

P(a₁, a₂, …, aₘ), that the sequence (a₁, a₂, …, aₘ) has our property in that

*) Here is the title and exact “address” of the article: Yu. V. Matiyasevich. Diophantine property of the listed sets, DAN SSSR, 1970, volume 191, no. 2.

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only if there are numbers $x_1, x_2, \dots, x_n$ for which $P(a_1, a_2, \dots, a_m, x_1, x_2, \dots, x_n) = 0$.

Matiyasvich’s main achievement is precisely the proof of the Diophantine property of a certain special property of pairs of numbers $(a, b)$.

Many of you are familiar with Fibonacci numbers.

$$\psi_0 = 0, \quad \psi_1 = 1,$$
$$\psi_2 = \psi_0 + \psi_1 = 1,$$
$$\psi_3 = \psi_1 + \psi_2 = 2,$$
$$\psi_4 = \psi_2 + \psi_3 = 3,$$
$$\psi_5 = \psi_3 + \psi_4 = 5,$$

an infinite sequence of which is determined by the recurrent formula

$$\psi_{n+1} = \psi_{n-1} + \psi_n.$$

The property of the pair of numbers $(a, b)$ that Matiyasviči deals with is:

“$b$ is the Fibonacci number

with number $2a$”, i.e. $b = \psi_{2a}$.

It is clear that pairs have this property

$(0,0), (1,1), (2,3), (3,8)$

etc. The statement that the Matiyasvich property is Diophantine means that there is a polynomial

$$Q(a, b, x_1, x_2, \dots, x_n)$$

such that $b$ will be a Fibonacci number with number $2a$ if and only if there exist numbers $x_1, x_2, \dots, x_n$ such that

$$Q(a, b, x_1, x_2, \dots, x_n) = 0.$$

The polynomial $Q$ with integer coefficients could, in principle, be written out explicitly. Matiyasvich does not do this and does not even indicate what the number $n$ of additional variables is.

As for the proof of the formulated theorem, then, as already said, it is quite elementary. Matiyasvich refers to one lemma proven in N. N. Vorobyov’s book “Fibonacci Numbers,” intended for schoolchildren, and one result proven in a more scientific book by A. I. Maltsev, but also elementary. In Matiyasvich’s work itself, nineteen lemmas are given without proof, but the proof of each of them separately is no more difficult than typical problems for schoolchildren. The difficulty was only to find that particular combination of elementary proposals that leads to the final goal.

2. Why is all this so important?

If the polynomial $Q$ Matiyasvich was really needed by a mathematician to carry out any calculations, then it is believed that Matiyasvich would not be too lazy to write it down explicitly. But in reality, the very appeal to the Fibonacci numbers is eurysimable for sequentially writing out all pairs possessing this property

$(0,0), (1,1), (2,3), (3,8), (4,21), (5,55), \dots$

Since the polynomial of these pairs is infinite, the process of writing them out will never end. But we have a rule according to which our pairs can be written out one after another with magnification in that

1. Only pairs with the Matiyasvich property will be written out,
2. any pair possessing the Matiyasvich property will sooner or later be discharged.

A property (in more scientific language, a “predicate”) of a finite sequence of numbers $(a_1, a_2, \dots, a_m)$ is called an enumeration if there is a rule that allows, through the mechanical application of this rule, to obtain sequences one after another that consistently have a given property, and moreover, in such a way that any sequence

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A property with this property will sooner or later be obtained.

From a modern point of view, the most significant consequence of Matiyasvich’s theorem on Fibonacci numbers is

Theorem 1. Any enumeration property of a finite sequence of numbers is Diophantine.

Before 1961, this theorem would have seemed extremely surprising. But some weaker results by Davis, Putnam and Robinson, proven in 1961, made such a hypothesis no longer so incredible. Considerable effort was spent on proving her mathematics. Matiyasvich took advantage of some, so to speak, “reductions” of the problem to more special problems. But the idea of ​​reducing everything to the properties of Fibonacci numbers was also unexpected in the situation that had developed by 1970.

Thus, Matiyasvich’s major success required combining an understanding of the big problems of modern mathematics and erudition in his field with the art of finding unexpected, completely elementary ways to solve special problems.

3. In what sense did Matiyasvich solve Hilbert’s tenth problem?

In Hilbert’s tenth problem*) he solved equations of the form

$$P(x_1, x_2, \ldots, x_n) = 0,$$

where $P$ is a polynomial with integer coefficients. The degree of such an equation is called the degree of the polynomial $P$. When the question is asked about finding their solution in integers, these equations are called “Diophantine”. For example, the equation

$$x^2 + y^2 = 1$$

has even solutions in integers

$$x = \pm 1, \quad y = \pm 1,$$

*) More details about Hilbert’s problems can be read in the book D. Hilbert, “Mathematical Problems”, “Science”, 1969.

and the equation

$$x^2 + y^2 = 3$$

has no solutions in integers.

Now let’s give the floor to Hilbert himself:

“Let an arbitrary diophantine equation with an arbitrary number of unknowns be given…; We need to indicate a general method, the following procedure could be used to find out in a finite number of steps whether we have a given equation of solution in integers … or not.”

Hilbert was apparently convinced that the general method he was looking for existed, and the only question was to find it. A lot of effort was spent to find this method, but no encouraging results were obtained.

The problem of integer solutions to an arbitrary equation can easily be reduced to the problem of natural (integer non-negative) solutions. Further, it is not at all difficult to show that it is sufficient to restrict ourselves to Diophantine equations of degree no higher than four. For diophantine equations of degree no higher than the second, the sought general method was found, but equations of the third degree were no longer fed into any numbers.

In connection with the failures in this direction, the suspicion arose that the general method that Hilbert’s formulation was supposed to find simply did not exist! A similar situation has developed in a number of other problems of a similar nature. However, it is one thing to find the required general method - here it was enough just to present this method and directly verify that it satisfies the conditions put forward by Hilbert.

But in order to prove the non-existence of a certain method for solving a series of problems, it was necessary to give an exact definition of what this general method is, how and by what means it can be implemented.

At the beginning of the thirties, the corresponding definitions were expanded

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nerds in the works of the American scientist Church and the English scientist Tyorig. These definitions laid the foundation for the theory of algorithms. Looking back on the path traveled, mathematicians should be grateful to Hilbert’s ten problems for the fact that it served as one of the incentives for the creation of this theory.

Within the framework of the theory of algorithms, a precise definition of the concept “enumerable property” was obtained, which we used in the previous paragraph.

It turned out that from the Diophantine property of any enumerated property of a finite number sequence it follows

Theorem 2. It is impossible to have a general method (algorithm) that allows us to determine for any given Diophantine control whether it has solutions in integers or not.

Matiyasevich from his theorem on the Diophantine property of a special property of a pair of numbers (a, b)

$$b = \psi_{2a}$$

derived a proof of Theorem 1, from which, as was already known, Theorem 2 follows, i.e., a negative solution to Hilbert’s ten problems.

For a final understanding of everything that has been said, you only need a clear knowledge of what an “algorithm” is (a general method for solving an infinite series of problems) and what an “enumerable property” (an enumerative predicate) is. But this is already a much more difficult topic.

4. Definitions and facts of the theory of algorithms

5. Among the functions defined on the set N of natural numbers and taking natural values, the class of primitive recursive functions is distinguished as follows.

Consider the function S(n) n+1 and q(n)=n-p, where p²≤n<(p+1)². From the functions f(n) and q(n) it is cut to form the function h₁(n)=f(n)+q(n) (addition operation), $h_2(n)=f(q(n))$ (substitution operation), and from one function f(n) - a function defined by the equalities h₃(0)=0, h₃(n+1)=f(h₃(n)) (ineration operation). The class of primitive recursive functions consists of the functions S, q and all those functions that can be obtained from them by a finite number of additions, substitutions and iterations.

For example Let us show that the function φ(n)=2n and φ(n)=2n+1 are primitively recursive. Let us first apply the ineration operation to the function S(n)=n+1. Then h(0)=0, h(n+1)=S(h(n))=h(n)+1. Therefore, h(n)=n and φ(n) is obtained by applying the addition operation to two functions, each of which is equal to h(n): φ(n)=h(n)+h(n)=n+n=2n. Finally, the function φ(n) is obtained by a substitution operation applied to the functions S(n)=n+1 and φ(n)=2n: φ(n)=S(φ(n))=S(2n)=2n+1.

2. A set M of natural numbers is called enumerative if it coincides with the set of values ​​of some primitive recursive function.

For example, the set of even numbers is enumerable, since it is the set of values ​​of the primitive recursive function φ(n)=2n. The set of odd numbers coincides with the set of values ​​of the primitive recursive function φ(n)=2n+1, and is also enumerable.

3. A set M is called decidable if it is enumerable together with its complement N - M (where N is the set of all natural numbers).

In particular, a solvable set of numbers, since it is enumerable along with its complement (the set of odd numbers).

4. There are enumerable but undecidable sets.

It is possible to give a specific example of such a set, but the corresponding construction is technically too complex to be carried out within the framework of this comment.

Now let this be a set of M natural numbers. One may ask whether there is a general method that, given every positive integer n, determines in a finite number of steps whether that n belongs to the set M or not. The main position of the theory of algorithms (Church’s theory) states that any method (algorithm) exists, model and model, when the set M is solvable.

To obtain a negative solution to Hilbert’s agent problem, it was sufficient to prove that each enumerated set is diffant, i.e., for each enumerated set $M$ one has to construct a diffantian equation $P(y, x_1, \dots, x_h) = 0$ that would have natural solutions $x_1, \dots, x_h$ for all $y$ belonging to $M$, and only for such $y$.

In fact, if this could be done, then, taking as $M$ an unsolvable set (there are such among those listed, see item 4), we would obtain that for the corresponding equation $P(y, x_1, \dots, x_h) = 0$ there is no general method (algorithm) that, for each natural number $y$, would give an answer to the question about the existence of natural solutions for this equation. After all, if this method existed, then it would be possible to find out in a finite number of steps whether the equation $P(0, x_1, \dots, x_h) = 0$ has a solution (i.e., does the number 1 belong to the set $M$), does the equation $P(1, x_1, \dots, x_h) = 0$ have a solution (i.e., does the number 1 belong to the set $M$), etc. It would turn out that there is a general method that, for every natural $y$ determines in a finite number of steps whether this $y$ belongs to the set $M$ or not. Then, by Church’s test $M$, the question of choosing this set would be solvable.

5. What was done and what Matiyasevich did.

In the fifties, a group of American mathematicians (R. Robinson, H. Putnam, M. Davis, J. Robinson) received a number of significant discoveries.

current results in Japanese proofs of diffantency of enumerable sets. (The conjecture that enumerable sets are diffant was advanced by Martin Davis.)

American scientists value the significant conditions for being able to reduce the problem to proving the diffantency of the relation $y = z^u$. Julia Robinson even went a little further, showing that it is enough to construct a specific equation $R(u, v, x_1, \dots, x_h) = 0$ that does not admit a solution with $v > u^u$, but for each $n$ it has a solution with $v > u^n$. It was Yu. V. Matiyasevich who was able to construct this kind of equation and who proposed a completely elementary, but extremely witty and original construction. At the same time, we had to solve a problem that was unusual for traditional number theory. Indeed, in number theory, for a given equation, as a rule, the properties of its solutions are studied, but here, on the contrary, having given certain properties of the solutions, it was necessary to search for the required equation.

Turning to the consideration of the Fibonacci sequence, Matiyasevich noticed that if $u$ is taken to be half the number of the even term of the sequence, and $v$ is the term itself, then the inequality $v > u^u$ will always be false, and for any $n$ one can find such a cell term of the sequence.

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It is certain that the inequality $v > u^n$ will be true. This circumstance can be constructed in the table given on page 43. (It highlights those cells in which the numbers $u^k$ are less than the corresponding numbers $v$).

After the indicated property was verified, Matiyasevich “corrected” the Fibonacci sequence, removing odd terms from it. Namely, he considered the sequence given by the corresponding ones: $\varphi_0 = 0, \varphi_1 = 1, \varphi_{n+1} = 3\varphi_n - \varphi_{n-1}$. The result is exactly a sequence of four members of the first sequence: $\varphi_0 = 0, \varphi_1 = 1, \varphi_2 = 3, \varphi_3 = 8, \varphi_4 = 21, \varphi_5 = 55, \varphi_6 = 144, \varphi_7 = 377$, etc. By taking $u$ as the number of the sequence member, and $v$ as the sequence member with number $u$, i.e. $\varphi_u$, we find that the required property is again satisfied for $u$ and $v$. Terer only had to construct the equation $P(u, v, x_1, \dots, x_k) = 0$, which would have a natural solution if and only if $v = \varphi_u$ - the rest was based on the result of Julia Robinson! To do this, it was enough to construct a system of diffant equations $P_1 = 0, \dots, P_n = 0$ in variables $u, v, x_1, \dots, x_k$, which has a solution if and only if $v = \varphi_u$ - this system has exactly the same solutions as the only equation $P_1^2 + \dots + P_n^2 = 0$.

This is the form in which Matiyasevich obtained the required system of equations:

1. $u + (a - 1) = v$,
2. $v + b = l$,
3. $l^2 - lk - k^2 = 1$,
4. $g^2 - gh - h^2 = 1$,
5. $l^2c = g$,
6. $ld = r - 2$,
7. $(2h + g)e = r - 3$,
8. $x^2 - rxy + y^2 = 1$,
9. $lp = x - u$,
10. $(2h + g)q = x - v$.

To prove that the existence of a natural solution to this system is equivalent to the corresponding $v = \varphi_u$, Matiyasevich used the following characteristic property of the sequence $\{\varphi_k\}$:

a pair of conditions $x < y$ and $x^2 - 3xy + y^2 = 1$ is equivalent to the existence of a $k$ such that $x = \varphi_k$ and $y = \varphi_{k+1}$.

We recommend that the interested reader try his hand at proving this equivalence.

Let us point out in conclusion that the found solution to the ten problems has a number of interesting consequences, of which, perhaps, the most effective is the following: you can specify a specific linear polynomial, the set of positive values ​​of which coincides with the set of all prime numbers!

Mathematical science has recorded a server and honorable achievement.

Task 1

1. Show that for any $k$ the numbers $\varphi_k$ and $\varphi_{k+1}$ are coprime.
2. Prove compliance: $\varphi_{k+1} = \varphi_k + 1$.
3. Show that the diffant equation $P(y, x_1, \dots, x_k) = 0$ has a solution for positive numbers if and only if the equation has such a solution

$$Z = (1 - (P(z, x_1, \dots, x_k))^2) - y.$$

Due to the fault of the printing house, in «Kvant» No. 5 (p. 64, task 1226), the formula for dividing numbers was typed incorrectly. It should look like this:

$$\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}$$

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In our century, physicists have become interested in a great thing - a central accelerator, a magnet, a field of hundreds of thousands of fields, a quantum electronics device. Only by building a huge camera to observe processes in the microcosm can physicists penetrate data into the essence of natural phenomena. But not so long ago, at the beginning of our century, the physicist himself built small installations, shrunk on his table and made great discoveries. The atomic nucleus, radioactivity, RENTen rays, radio waves - all this was discovered in the 19th century with the help of simple means that were at the disposal of the natural body. What if the data is still missing? It’s hard to believe now that in the 18th century even the royal doctors could not communicate with each other about the king’s pulse rate, since their clocks ran differently. Nobody measured blood pressure or even just body temperature. The appearance of the thermometer caused no less enthusiasm then than the discovery of the laser today. This is how excited Reaumur wrote about this invention in 1730.

Y. A. Smorodinsky

“ON THE CONSTRUCTION OF THERMOMETERS”, 1730

Reaumur

The thermometer, without a doubt, is one of the wonderful inventions of modern physics, which in turn contains a lot of significant information. He provided us with more interesting knowledge that would have been inaccessible without his help. How could we, in many cases, without a thermometer, know that a liquid (some) mixed together is heating up? Without a thermometer, we would never have discovered what cooling occurs during the dissolution of known salts, and also in what cases this cooling is most pronounced. Nor could we know that one piece of ice could be colder than another. Nor could we discover that boiling water has a temperature above which water cannot be heated at all (at normal pressure).

All physicists know that with a thermometer in your hands you can carry out countless experiments. But this device requires more than one physicist; Its use is not limited to laboratories - we usually know from a thermometer what the air temperature is…

Doix por (i.e. before 1730) thermometers were almost never used to determine the coldest and hottest in different climates - questions of the greater half and intersas. Meanwhile, in this way it is easy to find out how many degrees of heat or cold a person can withstand. It would be equally important to know at what temperatures it is necessary to grow those plants and trees that could acclimatize here.

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All mathematics problems in this issue are taken from those presented at the last Moscow Mathematical Olympiad; some wording should be changed.

F36. A drop of liquid lies on the flat side of a hemispherical glass plate. Show how the refractive index of a liquid can be determined from observing total internal reflection. The refractive index of the stack is also unknown and needs to be determined.

47

F37. In two vertically located cylinders, the cross-sectional areas of which are $S_1$ and $S_2$, there are two irreversible pistons connected by a thin irreversible thread of length $l$ (Fig. 1). The space between the pistons is filled with water. Find the tension of the thread if the ends of the vessels are open to the atmosphere. The density of water is $\rho$.

F38. The wide elbow of a U-shaped mercury manometer has a larger diameter than the narrow one. To which elbow should a pressure measurement report scale be attached to ensure greater measurement accuracy?

F39. A planet of radius $R$ and mass $M$ is surrounded by an equally dense atmosphere consisting of gas with molecular weight $\mu$. What is the temperature of the atmosphere on the surface of the planet if the altitude of the atmosphere is $H$?

Physics Olympiad of Moscow State University, 1967

F40. At what potential difference between the electrodes does a neon light bulb light up if the ionization energy of neon is $I=21,5$ eV, and the average distance between two successive collisions of the electrode with gas atoms is 0.4 mc? The electrodes look like large plates located at a distance of $d=3$ mm from each other.

F41. How many times is the security on a moonlit night during a full moon less than on a sunny day, with the same height of the Moon and Sun above the horizon? Assume that the illuminated resulting harrier uniformly scatters light into space. The radius of the Moon is taken to be 2000 km, and the distance from the Moon to the Earth is 400,000 km.

M31. A square sheet of paper is cut straight into two parts. One of the resulting parts is again cut into two parts, and this is done many times. What is the minimum number of cuts that must be made so that among the resulting parts there will be an equal number of dimensions above?

I. Bernishtein

M32. All cells of the table $100 \times 100$ contain pluses. It is allowed to simultaneously change the signs in all cells of one row or one column. Is it possible, by continuing the operations several times, to obtain a table with exactly 1970 minutes?

A. Zelevinsky

M33. There is a natural number $n>1000$. Let’s take the remainders from dividing the number $2^n$ by the numbers 1, 2 3, . . . , $n$ and find the sum of all these remainders. Prove that this sum is greater than $2n$.

A. Kuinirenko

M34. Prove that if a natural number is divisible by 10 101 010 101, then in its decimal notation at least six digits are different from zero.

A. Tolpieo

M35. A certain 19-facet is described near a sphere of radius 10. Prove that you will find two points on its surface, the distance between which is greater than 21.

A. Kuinirenko

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From this issue we will begin printing the solution to the Quantum Problem Book.
Most of the problems placed in this section of the journal are quite difficult and a discussion of some of them should be devoted to a larger statue. Many of them give rise to a conversation about a whole class of problems, about methods of approaching certain phenomena, about the use of various phenomena in practice. Sometimes we will have this conversation on the pages of our magazine. But often the reader himself should think about how to generalize the given solution, or how to simplify it, about where the phenomenon mentioned in the assignment is applied, etc.

The seventh issue was prepared for printing at the end of March, so, naturally, letters with decisions that we received later are not taken into account here. In the future, it is expected to publish the solution approximately six months after the publication of the problems. At the same time, we successfully got acquainted with the latest solutions “on time” - no later than a month and a half after the release of the corresponding issue. Of course, we will not be able to respond to all readers who sent us letters. The best solutions will be published on the pages of the magazine. Those who came up with a solution that is very different from the published one will have to figure out for themselves whether it is correct or not, and if not, what is the mistake.

And one last thing. Coming up with a problem is generally more difficult than solving it. It’s more difficult, but it’s also more interesting. Try to come up with problems and send them to us. We will publish the best problems in the Quantum Problem Book.

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M1. In the country of Anchuria, where President Miraflores rules, the time for new presidential elections is approaching. There are approximately 20 million voters in the country, of which only one percent (the regular army of Anchuria) supports Miraflores. Miraflores naturally wants to be elected, but on the other hand he wants the elections to appear democratic. What Miraflores calls “democratic voting” is this: all voters are divided into several equal groups, then each of these groups is raised by a number of equal groups, then these last groups are again divided into equal groups, etc.; in the smallest groups, representatives of the elector group are elected, then the elector submits for voting in an even larger group, etc.; finally, representatives of the largest groups choose a representative for the groups as he wishes, and instructs his parties how to vote. Will he be able to organize “democratic elections” in such a way that he will be elected president? (If the votes are equal, the opposition wins.)

Answer. Yes, we can.

First of all, let’s figure out how a candidate with a minority vote can win in a “multi-stage” election. (By the way, many capitalist countries vote under this system.) The simplest example of such a situation is depicted in figure 1: here nine voters - four “black” and five “blue” - are divided into three groups of three voters so that in two groups the blacks win, and therefore, as a result of this “two-stage election”, the black candidate will be elected, although the number of his side is only $\frac{4}{5}$ of the total number of voters. (It is not difficult to report that in a two-stage election with a large number of voters, the percentage of votes sufficient to win may be even less, but everything is still completed above 25%.) It is clear that in a three-stage election system this percentage can be made even lower. For example, if we replace by figure 1 each voter in a group of one hundred people, so that in the voting group all of these voters are voting, and in the black group there are 51 blacks and 49 blues, then we get an example of a situation where blacks make up only $\frac{4}{9} \cdot \frac{51}{100} = \frac{17}{75}$ of the total number of voters and nevertheless win.

After these preliminary messages, the solution to the problem came.

Let’s divide all voters into 5 groups of 4 million each so that two groups will consist entirely of opponents of Mirafloresa (let’s call these groups “blue”, and the other three “black”). Each of

50

These “first rank” groups will again be divided into 5 groups of second rank, and of the five groups that make up the black group of the first rank, three are black, etc., as indicated in the table.

It is clear that with such a division, for Miraflores to win, it is enough for people to vote for him

$$\frac{3^{11}}{2^8 \cdot 5^7} = \frac{177 \cdot 147}{20 \cdot 000 \cdot 000} < \frac{1}{100}$$

part of all voters. Thus, since the army constitutes $\frac{1}{100}$ part of the voters and supports Miraflores, he will be able to win.

Some readers have tried to solve the following, more general question: what is the smallest number of supporters Miraflores must have to win a “democratic election” if the total number of voters is $N$? Of course, the answer to this question depends not so much on the size of the number $N$, but on how it is factored. If, say, $N$ is a prime number, then voters cannot be divided into equal groups in any way (except for the trivial method: $N$ groups of one person each) and to win you need to have a simple greater instrument. We’ll show you how to answer this question for any $N$.

Let us consider such a division of $N$ people into groups (1st, 2nd, etc. rank), in which Miraflores wins and the number of his side is the smallest possible. Obviously, it can be assumed that in the groups voting against him there is not a single one on his side and that all black groups of the same rank are divided exactly the same. It is convenient to use communication $[1]$ - “integer part” ($[x]$ - the most integer number, not exceeding $x$; for example, $[2] = 2$, $[3\frac{1}{2}] = 3$). It is clear that if a non-orthodox black group consists of $k$ groups of the next rank, then there must be at least $[\frac{k}{2}] + 1$ blacks among them. Let each group of $r$ rank ($r = 1, 2, \dots, m-1$) be divided into $k_r$ groups of lesser rank, and let the groups of the last $m$ rank consist of 1 person. Then to win you must have at least

$$B = \left( \left[ \frac{k_1}{2} \right] + 1 \right) \left( \left[ \frac{k_2}{2} \right] + 1 \right) \dots \left( \left[ \frac{k_m}{2} \right] + 1 \right)$$

black voices. Our task boiled down to this: decompose the given $N$ into a product of maximals $k_1, k_2, \dots, k_m$ so that the product $B$ is minimal.

Let

$$N = k_1 k_2 \dots k_m$$

• such decomposition. As the next lemma shows, we can propose that in this expansion there are no factors $k$ of the form $k = pq$, where $p$ and $q$ are greater than 2 (if there are such, we can check the further expansion without increasing $B$).

Lemma.

$$\left( \left[ \frac{p}{2} \right] + 1 \right) \left( \left[ \frac{q}{2} \right] + 1 \right) \leq \left[ \frac{pq}{2} \right] + 1$$

for all integers $p$ and $q$ greater than 2.

Proof. If $p$ and $q$ are both even, then the inequality can be rewritten as:

$$\left( \frac{p}{2} + 1 \right) \left( \frac{q}{2} + 1 \right) \leq \frac{pq}{2} + 1,$$

$(p + 2) (q + 2) \leq 2pq + 4,$

$pq - 2q - 2p + 4 \geq 4, \quad (p - 2)(q - 2) \geq 4,$

which is obviously true for $p > 2$ and $q > 2$.

In the case when one of the numbers $p, q$, for example $p$, is even and the other is odd, we have:

$$\left( \frac{p}{2} + 1 \right) \left( \frac{q}{2} + 1 \right) \leq \frac{pq}{2} + 1,$$

$(p + 2) (q + 1) \leq 2pq + 4,$

$pq - 2q - p + 2 \geq 0, \quad (p - 2)(q - 2) \geq 0.$

The case where $p$ and $q$ are odd is split similarly. The lemma is proven. From here it immediately follows that odd $N$ must be factored “to the end” into prime factors. All that remains is to deal with the deuces.

We can assume that the expansion contains no factors of the form $2q$, where $q$ is odd, since $2$ is of rank

$$2 \left( \left[ \frac{q}{2} \right] + 1 \right) = \left[ \frac{2q}{2} \right] + 1.$$

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necks through the centers of the circles and the center of the sphere. Indeed, both circles are symmetrical with respect to the plane $P$, and if they had a common point on one side of the plane $P$, they would have to have a symmetrical common point on the other side $P$, but they have only one common point. If these circles have the same radius $r$, then the distance between their “centers” is equal to $2r$, because on the great circle circle resulting from the intersection of the sphere and the plane $P$ (Fig. 2, $\theta$), the diameters of our circles (black segments) and the segment connecting their “centers” (red) contract equal arcs.

Let $A_0, A_1, A_2, \dots, A_n$ be the “centers” of circles $\gamma_0, \gamma_1, \dots, \gamma_n$, which are discussed in the conditions of the problem. Then $A_0A_1 = A_0A_2 = \dots = A_0A_n = A_1A_2 = A_2A_3 = \dots = A_nA_12r$, in other words, $A_0A_1A_2 \dots A_n$ is a regular $n$-gonal paramid inscribed in a given sphere of radius 1 with vertex $A_0$, all of whose lateral faces are equilateral triangles with sides equal to $2r$. So, it is enough to construct a paramid for which these conditions are met, then the points $A_0, A_1, \dots, A_n$ will be certain circles of radius $r$ with “centers” $A_0, A_1, \dots, A_n$, which obviously satisfy the conditions of the problem.

Since the sum of the plane angles of a convex $n$-faceted angle with vertex $A_0$ is less than $360^\circ$:
$$n \cdot 60^\circ = A_1A_0A_2 + A_2A_0A_3 + \dots + A_nA_0A_1 < 360^\circ,$$
then $n < 6$. For $n = 3, 4$ and 5 it is not difficult to construct the necessary paramids.

Let $O$ be the center of the sphere. The height of the paramid $h$ and the length of its edges $2r$ are found from the following considerations: the radius $KA_1$ of the base of the paramid is the leg $\Delta A_0KA_1$ and the side $\Delta A_1KA_2$, where $\Delta A_1KA_2 = 2\pi/n$ (Fig. 3, $a, \delta$),

$$\sqrt{4r^2 - h^2} \sin \frac{\pi}{n} = r.$$

From $\Delta A_0OA_1$ we have $r = \frac{h}{2r}$.

Hence $h = 2r^2$, $r = \sqrt{1 - \frac{1}{4 \sin^2 \frac{\pi}{n}}}$.

Thus,

at $n = 3$: $r = \sqrt{\frac{2}{3}} \left( \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \right)$;

at $n = 4$: $r = \sqrt{\frac{1}{2}} \left( \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} \right)$;

at $n = 5$: $r = \sqrt{\frac{5 - \sqrt{5}}{10}}$

(the formula $\sin \frac{\pi}{5} = \frac{\sqrt{10 - 2\sqrt{5}}}{4}$ can be derived from figure 4, with the help of which regular decagons and regular heel-gons are constructed).

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Knowing $r$ and $h$, we can construct correct paramidas, which, due to the above relations, will satisfy all the necessary conditions: all faces are equilateral triangles with sides $2r$, the radius of the circumscribed sphere is 1.

The constructed paramids are closely related to regular polyhedral

kami, the faces of which are triangles. There are only three such polyhedra: tetrahedron, octahedron and ikocahedron (Fig. 5), and if from each of these polyhedra we cut off the “top” - all the faces adjacent to one vertex - then we get exactly the same three paramides that we were studying. You can read more about regular polyhedra and the construction of a pentagon in the excellent book by G. S. Kokter “Introduction to Geometry” (“Science”, 1966, Chapter 10 “Five Platonic Solids” and Chapter 11 “Golden Ratio and Phyllotaxis”).

A 10th grade student Serdey Makeev from Volokolamsk solved the problem somewhat differently (without using “centers on the sphere”).

Let us also note that the constraint $n \geq 3$ in the conditions of the problem could be completely replaced by $n \geq 2$. A corresponding arrangement of three circles $\gamma_0, \gamma_1, \gamma_2$ exists (Fig. 6; the “centers” of the circles are located at the vertices of a regular triangle-

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Fig. 6.

ka inscribed in a large circle), and to calculate $r$, oddly enough, the same formula that we proved for $3 \leq n \leq 5$ is suitable

$$r = \sqrt{1 - \frac{1}{4 \sin^2 \frac{\pi}{2}}} = \frac{\sqrt{3}}{2}.$$

M3. a) On figure 7, $a$ the plane is covered with squares of five colors. The centers of squares of the same color are located at the vertices of the square grid. With what number of colors is a similar filling of the plane possible?

6. On figure 7, $b$ the plane is covered with hexagons of those colors so that the centers of hexagons of the same color form the vertices of a lattice of identical regular triangles. With what number of colors is a similar filling possible?

Note. In the first problem, the number of colors can be one (all squares are the same color) or two (like on a chessboard). In the second problem you can easily find a pattern with one color and with three colors. It is advisable to give a complete solution to the problems, i.e., describe all colorings that satisfy the specified conditions. However, please send incomplete solutions if you find them interesting. Think, for example, is there a solution to the second problem with thirteen colors?

We have not yet received complete solutions to this problem.

6th grade student Serey Eliseev (Moscow) sent us a coloring of a hexagonal grid with thirteen colors, which we reproduce, as far as printing capabilities allow, on figure 8;

Fig. 7 a, b.

he states (without explanation) that the answer in general is:

In problem a), the required coloring is possible if the number of colors is equal to the sum of two squares of integers:

$$1 = 0^2 + 1^2; \quad 2 = 1^2 + 1^2; \quad 4 = 0^2 + 2^2; \quad 5 = 1^2 + 2^2; \quad 8 = 2^2 + 2^2; \quad 9 = 0^2 + 3^2; \quad 10 = 1^2 + 3^2; \quad 13 = 2^2 + 3^2; \quad 16 = 0^2 + 4^2$$, etc.

In problem b) the solution is possible with the number of colors equal to the sum of two numbers from the sequence 0, 1, 3, 6, 10, 15, . . . , $\frac{n(n+1)}{2}$, . . .

$$1 = 0 + 1; \quad 3 = 0 + 3; \quad 4 = 1 + 3; \quad 6 = 0 + 6; \quad 7 = 1 + 6; \quad 9 = 3 + 6$$, etc.

Try to find out if these answers are correct? We will return to this issue later, but for now let us pay attention to one point in the formulation of the problem (reader V. Gutenacher from Moscow was the first to notice it).

Let’s start with task a). It requires finding colorings in which “the centers of squares of the same color are located at the vertices of a square grid.” But the condition does not say that this “large” square grid should be the same for different colors. Look at figure 9. Here, too, the squares of each of the three colors form a grid, but for red the “step” (side of the square) of the grid is $\sqrt{2}$, and for white and green it is 2, and even the grid is surface-

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Fig. 8.

chickpeas relative to each other. But if the condition of the problem is understood literally - and this is exactly what should be done on a written exam or at an Olympiad - then the coloring of Figure 9 satisfies the condition of the problem. With this understanding, the answer to problem a) will be this: for any number of colors. Indeed, from any coloring with $n$ colors, one of which is white, it is easy to make a coloring with $(n+1)$ color: you need to color the white squares in a “checkerboard order” with white and the new $(n+1)$ red (this is how the coloring of Figure 9 was obtained from a checkerboard red-white coloring).

With this understanding, task 6) becomes very simple.

Answer. For any number of colors greater than 2. Indeed, coloring with two colors is impossible, since if out of three hexagons, each two of which have a common side, two are covered in the same color, then all hexagons must be colored with this color. coloring pages with 3 and 4 colors are easy to find, and then it’s not difficult to color white hexagons in three colors, increase the number of colors by 2, then by 2 more, etc.

Of course, you can get a lot of different coloring pages this way. Having any coloring $A(k)$ colors, highlighting in

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F1. Three communicating vessels with water, the centers of which are at the same distance $a$ from each other, are covered with pistons of the same thickness, made of the same materials. Vertical identical rods are attached to the pistons, which are hingedly connected to the rod $AB$. At what point on the rod can a load be attached to it so that in the equilibrium position the rod remains horizontal, if the masses of the rod and rods are negligible compared to the masses of the pistons and the load. The diameters of the vessels are $d_1$, $d_2$ and $d_3$, respectively.

Torsion $AB$ and the pistons are in equilibrium under the action of seven forces: force $Q$, acting from the load, the force of gravity acting on the pistons—these forces are proportional to the areas of the pistons: $P_1 = k_1 d_1^2$; $P_2 = -k_1 d_2^2$; $P_3 = k_1 d_3^2$ and three fluid pressure forces on the pistons: $N_1$, $N_2$, $N_3$ (Fig. 10). Since the liquid levels in the vessels are the same, the pressure on all pistons is equal, and the pressure force is related to the area of ​​the pistons: $N_1 : N_2 : N_3 = d_1^2 : d_2^2 : d_3^2$. Therefore $N_1 = k_2 d_1^2$, $N_2 = k_2 d_2^2$, $N_3 = k_2 d_3^2$.

The equal forces applied to each of the pistons are equal: $F_1 = N_1 - P_1 = (k_2 - k_1) d_1^2 = kd_1^2$, $F_2 = kd_2^2$, $F_3 = kd_3^2$.

Denoting by $x$ the distance from the middle of the rod to the point at which the load is attached, we write down the conditions of the rod: 1) the sum of all forces acting on the rod is equal to zero: $F_1 + F_2 + F_3 - Q = 0$ and 2) the sum of the moments of all forces acting on the rod is equal to zero relative to the middle of the rod: $F_1 a - F_3 a - Qx = 0$. Substituting the expressions for $F_1$, $F_2$ and $F_3$ into these equations, and then solving these equations together, we find that

$$x = \frac{d_1^2 - d_3^2}{d_1^2 + d_3^2 + d_3^2} a.$$

None of the readers who sent a solution to this problem took into account the masses of the pistons. But only two - G. Kalshin from Kirovobad and A. Karpenko from the town of Lisiansk, Voroshilgovred region - explained why this could be done. Since the pistons are made of the same materials and have the same thickness, the additional pressure on the body does not depend on the diameters of the pistons. The condition for equilibrium of the pistons is the equality of pressure on the body in all cylinders (since the water levels in the cylinders are the same). Additional pressures due to the pistons satisfy this condition automatically.

F2. Two balls with masses $m_1$ and $m_2$ lie on a horizontal plane, fastened together by a spring with a stiffness $c$. The plane is smooth. The balls are bent, releasing the spring, then they are simultaneously released. Determine the periods of the resulting oscillations of the balls.

The center of mass of the system should not move (or can move uniformly and rectilinearly), therefore the balls vibrate in antiphase with the same frequency, and their deviations $x_1$ and $x_2$ from the equilibrium position satisfy the corresponding pieces of the spring with lengths $l_1$ and $l_2$ ($l_1$ and $l_2$ are the distances from the balls to the center of mass of the system;

$$l_1 = l \frac{m_2}{m_1 + m_2}, \quad l_2 = l \frac{m_1}{m_1 + m_2}).$$

The elongation of the $1/q$th part of the spring is always $q$ times less than the elongation of the entire spring, i.e. the $1/q$th part of the spring has a stiffness $q$ times greater elongation than

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The stiffness of the entire spring. Therefore $c_1 = \frac{m_1 + m_2}{m_2} c$. It follows that the period of oscillation of the balls

$$T = 2\pi \sqrt{\frac{m_1 m_2}{(m_1 + m_2) c}}.$$

It is interesting to check the answer by taking some limiting case. This is what E. Tikhonov, a 10th grade student from school No. 20 from the city of Volgda, did. Let us assume that the mass of $m_2$ is very large: $m_2 \gg m_1$. Then a ball with mass $m_1$ should oscillate like this; as if the second ball were fixed, and $T = 2\pi \sqrt{\frac{m_1}{c}}$. Let’s check our formula

$$T = 2\pi \sqrt{\frac{m_1}{c \left(1 + \frac{m_1}{m_2}\right)}} \approx 2\pi \sqrt{\frac{m_1}{c}}.$$

F3. Two springs were made from two identical pieces of steel wiring. The diameter of the turns of one of them is $d$, the other $2d$. The first spring, under the action of the load, stretched by one tenth of its length. How much of its length will the second spring stretch under the same load?

The elongation of the spring is equal to $\Delta l = n \cdot 2d \cdot \sin \frac{\alpha}{2}$, where $n$ is the number of turns of the spring, and $\alpha$ is the angle at which adjacent turns of the spring unfold (Fig. 11). Since the extension of the spring is small, this angle is also small $\sin \frac{\alpha}{2} \approx \frac{\alpha}{2}$. Therefore $\Delta l = nd\alpha$.

Angle $\alpha$ is proportional to moment

forces $F$, which stretch the coil: $\alpha = F \cdot d$. The force $F$ is equal to the weight of the load suspended from the spring, and is the same in both cases, so $\Delta l \sim nd^2$.

The diameter of the coils of the second spring is twice as large, and the number of coils is twice as small, therefore, the abscylton elongation of the second spring is twice as large as that of the first. Thus, the second spring will stretch $\frac{2}{5}$ of its length.

Many who tried to solve this problem correctly found that the elongation of the second spring is twice as great as the first, but they forgot that the second spring is half as long as the first, therefore the relative elongation of the second spring is equal to $\frac{1}{5}$, as they found out, and $\frac{2}{5}$.

F4. The cylinder contains purified gas, but it is not known what kind. To raise the temperature of 1 keto gas by one degree at constant pressure, 958.4 J is required, and at constant volume - 704.6 J. What kind of gas is this?

When heating a gas at a constant volume of energy, the energy expended only goes towards changing the internal energy of the gas, and when heating at a constant pressure, it also goes towards doing work. Let us write down the law of conservation of energy for both cases:

$$mc_V \Delta t = \Delta W,$$

$$mc_p \Delta t = \Delta W + A.$$

Here $c_p$ is the specific heat of the gas at constant pressure (i.e., the amount of heat required to heat 1 kE of gas at constant pressure), $c_v$ is the specific heat of the gas at constant volume, $\Delta t$ is the change in temperature, $\Delta W$ is the change in the internal energy of the gas, $m$ is the mass of the gas, $A = p \Delta V$ is the perfect temperature during expansion gas work ($\Delta V$ - change in volume, $p$ - pressure).

Since when the temperature of a gas increases by the same number of degrees, the change in its internal energy is the same both when heated at a constant volume and when

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heating at constant pressure, we can write: $c_p m \Delta t = c_v m \Delta t + p \Delta V$. Using the equation of gas state (Clapeyron-Mendeleev equation), the work done by a gas can be expressed through the molecular mass of the gas $\mu$ and the gas constant $R$: $p \Delta V = \frac{m}{\mu} R \Delta t$. Substituting this correspondence into equation (1), we obtain: $c_p = c_v + \frac{R}{\mu}$, whence

$$\mu = \frac{R}{c_p - c_v} \approx 32,7 \text{ ккал/кмоль}.$$

The unknown gas is oxygen with a very small admixture of a heavier gas.

F5. There is an electrical target selected at figure. What will a voltmeter with a very high internal resistance show if it is connected to points $C$ and $D$? $U_{AB} = 518$.

The voltmeter will show the presence of voltage between points $E$ and $D$. In order to find it, it is convenient to first determine the voltage voltage between points $F$ and $B$. To do this, the circuit can be rewritten as shown in figure 12.

$$R_0 = 3 \text{ ом}, R'_0 = \frac{3}{4} \text{ ом},$$

$$U_{FB} = U_{AB} \frac{R'_0}{R_0^1 + R_1} = \frac{3}{11} U_{AB}.$$

Now, in the same way, find what

$$U_{ED} = \frac{R_3}{R_2 + R_3} U_{FD} = \frac{1}{3} U_{FD} = \frac{1}{11} U_{AB} = 4,6 \text{ е}.$$

This problem was correctly solved by almost all the readers who sent us letters.

F6. Find the value of a grounded metal ball of radius $r$ if there is a point charge $q$ at a distance $R$ from its center.

The ball’s potential must be zero. The field potential at the center of the ball, equal, of course, to the potential of the ball, is the sum of the field potential of the point charge $q$ and the field created by the charge $Q$ of the ball. The charge $Q$ is not distributed uniformly over the ball, but if the ball is divided into small sections with charges $\Delta q$, then the potential of the field created by the charge of the ball in the center can be expressed as the total potential of the fields of the exact charges $\Delta q$. Thus, we can write for the center of the ball:

$$\Delta \varphi = \sum \frac{\Delta q}{r} = \frac{1}{r} \sum \Delta q = \frac{Q}{r} + \frac{q}{R} = 0.$$

Hence $Q = -\frac{r}{R} q$.

Many readers received the correct answer, but we received very few correct solutions. Many readers have used the formula for the capacitance of a capacitor $C = \frac{\Delta Q}{\Delta \varphi}$, assuming the capacitance of the ball to be $r$. But this is true only for an isolated ball with a uniformly distributed charge.

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In this issue of Quaent we
We are limiting the publication of the most interesting
tasks from the new trial training
nickname for 5th grade.

1233. Out of 8 coins, one is counterfeit
      (easier). How to identify false
      sewing a coin by two weighers
      on a scale with two cups without a scale?

1234. Among 20 coins there is one
      fake, different in weight
      from real ones. Using three weights
      sewing on scales with cups without
      weights to detect a counterfeit coin
      and establish whether it is lighter or heavier than the real ones.

1235. I only have wall ones
      the hours that are left. I’m from-
      I’m going to see my friend, cha-
      whose sons walk faithfully, being
      he has some time and, having returned-
      When I get home, I set my watch correctly.
      How did I manage to do this if I didn’t first
      it was known how much time
      is the road long?

1236. For word page numbering
      6869 digits were required.

How many pages does the dictionary have?

1238. What is the divisor of the number
      3⁶·5⁴?

1239. When dividing the number N by 2 in
      the remainder is 1, and when dividing
      3 leaves a remainder of 2. Which one?
      the remainder will be obtained when dividing the number
      N for 6?

1240. If from the planned three-valued
      If you subtract 7 from a given number, then it will be divided by 7, but if you subtract 8 from it, then it will be divided by 8, and if you subtract 9, then it will be divided by 9. What number was intended?

1241. What digit is the number 66⁶⁶? 33³³? 7⁷?

1242. Prove that the number 7⁷⁷+1 is not divisible by 5.

1243. Prove that the sum of any two consecutive odd numbers is divisible by 4.

1244. Some six-digit numbers begin with the number 7. By dropping this number next and adding it to the right, we get a number that is 5 times less than the first initial one. Find the original number.

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1251. If you enter a zero between the digits of a two-digit number, then the resulting three-digit number will be 9 times larger than the first by 7, 11 and 13.

1252. In a six-digit number, the first digit coincides with the fourth, the second with the fifth and the third with the sixth. Prove that this number is divisible by 7, 11 and 13.

1253. Add one digit to the right and left of the number 43 so that the number is divisible by 45.

1254. Can the sum of four consecutive natural numbers be a prime number? What about the sum of three?

1255. Pupils A, B and C lined up in a 100m race. When A completed his run, B was 10m behind him. When B completed his run, C was 10m behind him. How many meters was C behind A when A finished his run?

1256. Show that any amount of money can be paid if the seller and buyer have only three- and five-kopeck coins.

1257. The father instructed his son to measure the length of the yard in steps. It was winter. To check, the father measured the same length of the yard with his steps. The father walked from the same place as the son and walked in the same direction, so that in some cases the tracks of father and son coincided. There are 61 footprints in the snow. What is the length of the yard if the son’s step is 0.54 m and the father’s step is 0.72 m?

1258. Prove theorems:

1. If the product of three integers is an odd number, then their sum is an odd number;
2. If the sum of two integers is an odd number, then their product is an even number.

1268. The edges of a right parallelepidea are expressed by natural numbers, and its volume is expressed by a prime number greater than 2. Prove that the sum of the lengths of two unequal edges is an even number.

1269. Can there exist a right paralleleped whose edge lengths are natural numbers and whose surface area is a prime number?

1270. Divide 7 apples into 12 people equally, without cutting any apple into 12 or more equal parts.

1271. Reduce fractions:
      $$\frac{37\ 373\ 737}{81\ 818\ 181}; \quad \frac{609\ 609\ 609}{205\ 205\ 205}.$$

1272. Find the meaning of the expression:
      $$\frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \frac{1}{5\cdot6} + \dots + \frac{1}{19\cdot20}.$$

1273. Prove that
      $$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \dots + \frac{1}{199} - \frac{1}{200} = \frac{1}{101} + \frac{1}{102} + \frac{1}{103} + \dots + \frac{1}{200}.$$

1274. Find the smallest natural number that, when divided by each of the fractions $\frac{28}{297}$ and $\frac{35}{396}$, produces integers.

1275. The train covers this distance in 10 numbers. If he increased his speed by 10 km/h, he would travel 8 numbers. Find the speed and distance.

1276. The car drove half the way at a speed of 50 km/h, and the second half at a speed of 30 km/h. At what average speed did the automatic travel the entire distance?

1277. From Gorky to Astrakhan the shuttle takes 5 days, and back 7 days. How many days does it take to travel from Gorky to Astrakhan?

1278. I drank $1/6$ a cup of black coffee and topped it up with milk. Then I drank $1/3$ cups and topped it up with milk again. Then I drank a cup of it. Did I drink more coffee or milk?

1279. The cook was preparing soup and putting little salt in it. I had to get fed up-

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cook the finished soup. The next time the cook recognized this mistake and added 2 times more salt to the same amount of soup than the first time. This also turned out to be not enough, but it was enough with half the amount of salt. How much salt did the cook put into the soup for the first time?

1283. Senya paid the canteen cash register for 3 dishes, and Sasha for 2 (all dishes were stacked equally). Kostya joined them at the table, and they were the second to eat these 5 dishes. When calculating, it turned out that Kostya had to pay his comrades 50 kopecks. How much money should Senya receive from this amount, and how much should Sasha?

1284. Front wheel purchases are automatically repaired after 25,000 km, and rear wheels after 15,000 km. When is it advisable to swap purchases so that they cost the same? (Assume that the front and rear purchases were replaced only once, although in practice drivers make this replacement more often.)

1285. Several chocolate lovers had breakfast together and had to pay 1 ruble. 75 kop. It turned out that the two did not have money with them, so each of the remaining ones had to pay 10 kopecks. more than everyone had to share. How many chocolate lovers had breakfast?

1286. The sum of three additions is equal to 1. The difference between the first and second additions is equal to the third. The sum of the first two dobies is 5 times greater than the third dobie. Find these dobies.

1287. The difference of two non-negative additions is $2/9$. The numeral of one dobi is 4 times greater than the other, and the denominator of this dobi is 3 times greater than the other. Find the dobie.

1288. Several natural numbers are given. Each of them was divided by the sum of all given numbers. What is the sum of all quotients?

1289. I thought of a three-digit number. If you make up all possible two-digit numbers from the digits of this number and then add them, then the required number is equal to the intended number. Find the target number.

1290. Mom bought apples for her children Vanya, Nina and Misha. The children had to divide the apples equally among themselves. Vanya came home first, counted the apples, took the third part and left. Then Nina came and, believing that she had arrived first, counted the remaining apples, took her share and left the same way. Finally, Misha came and took himself a third of the remaining apples. After all this, there are 8 apples left. How many apples does mother have for children?

1291. Two brothers would need to go to the railway station, which is located at a distance of 4 km from their house. In order not to be late for the train, there was only one option left - to ride a bicycle. But the older brothers have a certain weight. If you walk, you will be 10 minutes late. However, both brothers got to the station at the same time and 10 minutes before the train departed. Tell me, what did they do if walking is slower than riding a bicycle and if two people cannot ride a bicycle?

1292. What number has such a property that if you add an arbitrary number to it, then subtract the same arbitrary number from the product and divide the resulting difference by it, the result will be the same arbitrary number taken?

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WHAT do we know
And we don’t know
About gravity

Anyone who has read I. A. Efremov’s novel “The Andromeda Nebula” remembers the iron star - the “horror of astronomers”, which with its enormous attraction captures bodies flying past.

Even more amazing celestial lenses have to be the stars predicted by German astronomer Karl Schwarzschild. The gravitational forces near such stars are so strong that the light emitted by them cannot move beyond the immediate vicinity of the star. And a ray of light sent from afar to such a star will never be able to reach its surface. Some kind of strange gravitational mousetrap! But, despite all the necessity of such celestial bodies, the theory predicts their existence with a sufficient degree of rigor.

It would seem that the gravitational forces of the present are accustomed and harnessed, so that a person should have known everything about them that interests him a long time ago. It’s so simple: let’s put a stone on a scale and measure the force with which it is attracted to the Earth! But, having thought about the nature of such a phenomenon, we immediately find ourselves captive to unresolved issues.

In fact, why do any two bodies always attract each other and not repel under the influence of gravitational forces? Why can’t we isolate ourselves from the forces of gravity, just as a curtain separates us from the sun’s rays? What is the nature of the invisible
dim “springs” that attract bodies even very far from each other? There are many questions related to gravity that science has not yet been able to answer.

But we already know a lot. In 1687, Isaac Newton discovered the law of universal gravitation, which made it possible to understand how celestial bodies move. The next important step in understanding gravity belongs to Albert Einstein. In 1916, he created the so-called general theory of relativity, which is commonly called the theories of space, time and gravity. This theory forced scientists to shift their views to such important physical concepts as space and time. Newton believed that the properties of space and the passage of time do not depend on the surrounding bodies. Einstein proved that the heavier the body, the more the space near it is curved and the slower the time in its vicinity. Space, time and gravity turned out to be not independent of each other, but it is not easy to designate test. But its significance in modern physics is exceptionally great. Therefore, what new popular science book devoted to this topic (and there are very few such books) promises to arouse more interest.

We will talk here about some books that cover various problems of the modern theory of gravity.

A much simpler and more accessible presentation of the fundamentals of the theory of gravitation

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is contained in the popular book by M. Gardner, “The Theory of the Relative Length of Millions,” published in 1966 by Atomizdat. The fifth chapter talks about the main initial postulate of the theory - the equality of the gravitational and inertial masses of any body - confirmed with a great degree of accuracy by specially performed experiments. Chapter six describes the views of modern physics on gravitational forces as a result of changes in the geometric properties of space and time under the influence of gravitational masses. Chapters (7, 9 and 10) are related to various cosmic problems related to the theory of gravity. They consider different versions of the model of the Universe and discuss in detail questions about the so-called “red shift” - a change in the color of celestial objects under the influence of their movement, which occurs at enormous speeds.

A much more modern description of space problems can be found in a small brochure by academician V.L. Ginzburg, “How the Universe is Structured and How It Develops in Time,” published by the Znanie publishing house in 1968. (This brochure is based on three articles by the author in the magazine “Nuka and Life” Nos. 1, 2, 3 for 1968.)

First, the author talks about the first cosmological models of the Universe, which were proposed even before the creation of the general theory of the relative length of millions. Then, after a brief description of the main special and general theories of relative length, he examines new models associated with this theory. According to Einstein’s model, the Universe forms a closed three-dimensional sphere that does not change over time. In the model of the Soviet scientist A. A. Friedman, the Universe is either expanding or contracting. In Lemaître’s model, the Universe at some stage replaces its development, as if frozen in immobility, and then begins to rapidly expand again.

In L. Ginzburg provides interesting data on radio galaxies that are located at distances of about 5-8 billion light years from Earth. Their study tells us about the processes that took place at a time when there was not only the Earth, but, apparently, the Sun as well.

In 1969, the Nuka publishing house published P. Bergman’s popular science book “The Mystery of Gravitation.” The author of the book was an employee of Einstein for several years.

Bergman includes in his book all the main issues related to the modern theory of gravity. The book has three chapters: I. Newtonian physics and the special theory of the relative length of terrine. II. General theory of relative terrine length. III. Latest achievements. The book contains a commemorative account of the fundamental theories presented by his effective and unsolved problems on which physics has been working in our days.

Schwarzschild stars, gravitational collapse, the recent discovery of mysterious celestial objects “quasars”, cosmic radio emission telling us what the Universe looked like billions of years ago, and much more, no less interesting, describes the first cosmological models of the Universe, which were proposed even before the creation of the general theory of the relative length of millions. Then, after a brief description of the main special and general theories of relative terrine length, he considers new models associated with the modern theory of gravity. According to Einstein’s model, the Universe forms a closed three-dimensional sphere that does not change over time. In the model of the Soviet scientist A. A. Friedman, the Universe is either expanding or contracting. In Lemaître’s model, the Universe at some stage changes its development, as if STANCE in immobility, and then begins to expand rapidly again.

In 1969, the Nuka publishing house published P. Bergman’s popular science book “The Mystery of Gravitation.” The author of the book was an employee of Einstein for several years.

The author strives to help the reader. He puts all the basic calculations into a special mathematical supplement, and places at the end of the book a small explanatory dictionary of the most difficult and understandable terms. And yet, in some parts the book remains difficult to understand for high school students. True, there are many paragraphs in it that are accessible to a much less prepared reader. There are especially many of them in the third and final chapter of the book.

The books we have listed contain a whole world of new ideas and facts, future experiments, discovered and undiscovered phenomena and laws. And whoever touches this world will certainly receive great satisfaction.

V. A. Leshkovtsev

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ANSWERS, INSTRUCTIONS, SOLUTIONS

For the task “Only 2?”
(see «Kvant» No. 6)

From the equality $a + b = ab$ we find $a$:

$$a = \frac{b}{b - 1}.$$

Note that $b \neq 1$.

1. Let $a - \text{целое число}.$ Since the numbers $b$ and $b - 1$ are coprime (their difference is divided by their gcd, but it is equal to 1), then $b - 1 = 1$ or $b - 1 = -1$.

Hence $b_1 = 2$, $a_1 = 2$; $b_2 = 0$, $a_2 = 0$.

2. If $b$ is an arbitrary rational number not equal to 1, then using formula (*) we determine $a$.

3. If $ab = p$, then $p = \frac{b^2}{b - 1}$, whence

$$b^2 - pb + p = 0, \quad b = \frac{p}{2} \pm \sqrt{\frac{p^2}{4} - p},$$

$$b = \frac{p}{2} \pm \sqrt{\frac{p(p - 4)}{2}}.$$

a) using formula () for any $p \geq 4$ or $p \leq 0$ one finds the corresponding $b$ and using formula () $a$;

6. we need to find whether there is a number $p$ such that

$$p(p - 4) = \frac{q^2}{s^2},$$

(***)

where $\frac{q}{s}$ is an irreducible fraction. We solve equation (*) in integers $(p, q, s$ and $m$ - integers):

$$q^2 = s^2p(p - 4),$$ but from the irreducibility of the fraction it follows that $s = 1$.

$$p^2 - 4p - q^2 = 0, \quad p = 2 \pm \sqrt{4 + q^2},$$

$$4 + q^2 = m^2,$$

$$(m - q)(m + q) = 4 \cdot 1 = 2 \cdot 2.$$

The only solution in integers $m = \pm 2$, $q = 0$ leads to solution (1).

Two scenes from life

“Knowledgeable” tour guide

Guide: …Newton’s law applies to the planets, so they orbit around Coln. Kepler’s law also applies to them, but I won’t tell you about it, since it is complicated and you won’t understand anything… To fly away from the Earth, you need to have a second escape velocity. As you can see on the poster, it is equal to 11.2 km/sec, last year it was 11.4 km/sec… It is difficult for the astronauts. When they fly through the atmosphere, it puts a lot of pressure on the ship and the cosmonauts experience overload, but behind the atmosphere it’s even worse - the space of weightlessness begins there…

In one research institute

Graduate student: I believe that the roller sags in the bitumen due to the fact that its pressure increases the speed of the molecules, and they run away from under the roller.

Scientific supervisor: What exactly are you working on now?

Graduate student: Calculate the dependence of the speed of bitumen molecules on the pressure of the roller on them.

Scientific supervisor: Okay. What do you need to finish the job?

Graduate student: I would like a hydraulic press!

Scientific supervisor: Write an application. I think we’ll get it.

Three magic squares

65 4 54 64 6 53 70 3 50
24 44 55 23 43 57 20 40 63
34 75 14 36 74 13 33 80 10

67 9 47 69 8 46
26 37 60 25 39 59
30 77 16 29 76 18

72 2 49 66 5 52 68 7 48
19 42 62 22 45 56 27 38 58
32 79 12 35 73 15 28 78 17

A magic square is a square drawn into cells, each of which contains a number so that the sums of the numbers in any row, in any column and in both diagonals of the square are equal.

This 6x6 magic square can be made from the green 3x3 squares that are visible at figure. (Check!).

The task is to fill the cells of a 3x3 red square so that:
a) the red square was magical;
b) the entire large 9x9 square was also magical.

Crossword answer,
placed in no. 6.

Horizontal:
5. Scope. 6. Radian. 8. Sphere. 9. Point.
11. Tail. 13. Member. 14. Field. 16. Link.
19. Hatch. 21. Reception. 22. Polar. 23. Hadamard.
Vertical:

1. Carnot. 2. Mass. 3. Banh. 4. Gauss. 7. Libra.
2. Pendant. 12. Volt. 15. Wedge. 17. Coil.
3. Support. 19. Shmidt. 20. Ideal.

7