In the 7th grade mathematics course, we encounter for the first time equations with two variables, but they are studied only in the context of systems of equations with two unknowns. That is why a whole series of problems in which certain conditions are introduced on the coefficients of the equation that limit them fall out of sight. In addition, methods for solving problems like “Solve an equation in natural or integer numbers” are also ignored, although problems of this kind are found more and more often in the Unified State Examination materials and in entrance exams.

Which equation will be called an equation with two variables?

So, for example, the equations 5x + 2y = 10, x 2 + y 2 = 20, or xy = 12 are equations in two variables.

Consider the equation 2x – y = 1. It becomes true when x = 2 and y = 3, so this pair of variable values ​​is a solution to the equation in question.

Thus, the solution to any equation with two variables is a set of ordered pairs (x; y), values ​​of the variables that turn this equation into a true numerical equality.

An equation with two unknowns can:

A) have one solution. For example, the equation x 2 + 5y 2 = 0 has a unique solution (0; 0);

b) have multiple solutions. For example, (5 -|x|) 2 + (|y| – 2) 2 = 0 has 4 solutions: (5; 2), (-5; 2), (5; -2), (-5; - 2);

V) have no solutions. For example, the equation x 2 + y 2 + 1 = 0 has no solutions;

G) have infinitely many solutions. For example, x + y = 3. The solutions to this equation will be numbers whose sum is equal to 3. The set of solutions to this equation can be written in the form (k; 3 – k), where k is any real number.

The main methods for solving equations with two variables are methods based on factoring expressions, isolating a complete square, using the properties of a quadratic equation, limited expressions, and estimation methods. The equation is usually transformed into a form from which a system for finding the unknowns can be obtained.

Factorization

Example 1.

Solve the equation: xy – 2 = 2x – y.

Solution.

We group the terms for the purpose of factorization:

(xy + y) – (2x + 2) = 0. From each bracket we take out a common factor:

y(x + 1) – 2(x + 1) = 0;

(x + 1)(y – 2) = 0. We have:

y = 2, x – any real number or x = -1, y – any real number.

Thus, the answer is all pairs of the form (x; 2), x € R and (-1; y), y € R.

Equality of non-negative numbers to zero

Example 2.

Solve the equation: 9x 2 + 4y 2 + 13 = 12(x + y).

Solution.

Grouping:

(9x 2 – 12x + 4) + (4y 2 – 12y + 9) = 0. Now each bracket can be folded using the squared difference formula.

(3x – 2) 2 + (2y – 3) 2 = 0.

The sum of two non-negative expressions is zero only if 3x – 2 = 0 and 2y – 3 = 0.

This means x = 2/3 and y = 3/2.

Answer: (2/3; 3/2).

Estimation method

Example 3.

Solve the equation: (x 2 + 2x + 2)(y 2 – 4y + 6) = 2.

Solution.

In each bracket we select a complete square:

((x + 1) 2 + 1)((y – 2) 2 + 2) = 2. Let’s estimate the meaning of the expressions in parentheses.

(x + 1) 2 + 1 ≥ 1 and (y – 2) 2 + 2 ≥ 2, then the left side of the equation is always at least 2. Equality is possible if:

(x + 1) 2 + 1 = 1 and (y – 2) 2 + 2 = 2, which means x = -1, y = 2.

Answer: (-1; 2).

Let's get acquainted with another method for solving equations with two variables of the second degree. This method consists of treating the equation as square with respect to some variable.

Example 4.

Solve the equation: x 2 – 6x + y – 4√y + 13 = 0.

Solution.

Let's solve the equation as a quadratic equation for x. Let's find the discriminant:

D = 36 – 4(y – 4√y + 13) = -4y + 16√y – 16 = -4(√y – 2) 2 . The equation will have a solution only when D = 0, that is, if y = 4. We substitute the value of y into the original equation and find that x = 3.

Answer: (3; 4).

Often in equations with two unknowns they indicate restrictions on variables.

Example 5.

Solve the equation in whole numbers: x 2 + 5y 2 = 20x + 2.

Solution.

Let's rewrite the equation in the form x 2 = -5y 2 + 20x + 2. The right side of the resulting equation when divided by 5 gives a remainder of 2. Therefore, x 2 is not divisible by 5. But the square of a number not divisible by 5 gives a remainder of 1 or 4. Thus, equality is impossible and there are no solutions.

Answer: no roots.

Example 6.

Solve the equation: (x 2 – 4|x| + 5)(y 2 + 6y + 12) = 3.

Solution.

Let's highlight the complete squares in each bracket:

((|x| – 2) 2 + 1)((y + 3) 2 + 3) = 3. The left side of the equation is always greater than or equal to 3. Equality is possible provided |x| – 2 = 0 and y + 3 = 0. Thus, x = ± 2, y = -3.

Answer: (2; -3) and (-2; -3).

Example 7.

For every pair of negative integers (x;y) satisfying the equation
x 2 – 2xy + 2y 2 + 4y = 33, calculate the sum (x + y). Please indicate the smallest amount in your answer.

Solution.

Let's select complete squares:

(x 2 – 2xy + y 2) + (y 2 + 4y + 4) = 37;

(x – y) 2 + (y + 2) 2 = 37. Since x and y are integers, their squares are also integers. We get the sum of the squares of two integers equal to 37 if we add 1 + 36. Therefore:

(x – y) 2 = 36 and (y + 2) 2 = 1

(x – y) 2 = 1 and (y + 2) 2 = 36.

Solving these systems and taking into account that x and y are negative, we find solutions: (-7; -1), (-9; -3), (-7; -8), (-9; -8).

Answer: -17.

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Heinrich G.N. FMS No. 146, Perm

54 ≡ 6× 5 ≡ 2(mod 7),

55 ≡ 2× 5 ≡ 3(mod 7), 56 ≡ 3× 5 ≡ 1(mod 7).

Raising k to the power, we obtain 56k ≡ 1(mod 7) for any natural k. Therefore 5555 =56 × 92 × 53 ≡ 6 (mod7).

(Geometrically, this equality means that we go around the circle, starting from 5, ninety-two cycles and three more numbers). Thus, the number 222555 leaves a remainder of 6 when divided by 7.

Solving equations in integers.

Undoubtedly, one of the interesting topics in mathematics is the solution of Diophantine equations. This topic is studied in 8th, and then in 10th and 11th grades.

Any equation that needs to be solved in whole numbers is called a Diophantine equation. The simplest of them is an equation of the form ax+bу=c, where a, b and cÎ Z. The following theorem is used to solve this equation.

Theorem. The linear Diophantine equation ax+bу=c, where a, b and сО Z has a solution if and only if c is divisible by the gcd of the numbers a and b. If d=GCD (a, b), a=a1 d, b=b1 d, c=c1 d and (x0, y0) is a solution to the equation akh+bу=с, then all solutions are given by the formulas x=x0 +b1 t, y=y0 –a1 t, where t is an arbitrary integer.

1. Solve the equations in integers:

2. I considered the following problems with graduates in preparation for the Unified State Exam in mathematics on this topic.

1). Solve the equation in whole numbers: xy+3y+2x+6=13. Solution:

Let's factorize the left side of the equation. We get:

Since x,уО Z, we obtain a set of systems of equations:

Heinrich G.N.

FMS No. 146, Perm

Answer: (–2;11), (10; –1), (–4; –15), (–15, –3)

2). Solve the equation in natural numbers: 3x +4y =5z.

9). Find all pairs of natural numbers m and n for which the equality 3m +7=2n holds.

10). Find all triplets of natural numbers k, m and n for which the equality holds: 2∙k!=m! –2∙n! (1!=1, 2!=1∙2, 3!= 1∙2∙3, …n!= 1∙2∙3∙…∙n)

eleven). All terms of the finite sequence are natural numbers. Each member of this sequence, starting from the second, is either 14 times larger or 14 times smaller than the previous one. The sum of all terms of the sequence is 4321.

c) What is the largest number of terms the sequence can have? Solution:

a) Let a1 =x, then a2 = 14x or a1 =14x, then a2 =x. Then, by condition, a1 + a2 = 4321. We get: x + 14x = 4321, 15x = 4321, but 4321 is not a multiple of 15, which means there cannot be two terms in the sequence.

b) Let a1 =x, then a2 = 14x, a3 =x, or 14x+x+14x=4321, or x+14x+x=4321. 29x=4321, then x=149, 14x=2086. This means that the sequence can have three terms. In the second case, 16x=4321, but then x is not a natural number.

No answer; b) yes; c) 577.

12). All terms of the finite sequence are natural numbers. Each member of this sequence, starting with the second, or at 10; times more, or 10 times less than the previous one. The sum of all terms of the sequence is 1860.

a) Can a sequence have two terms? b) Can a sequence have three terms?

c) What is the largest number of terms the sequence can have?

Obviously, we can talk about the divisibility of integers and consider problems on this topic endlessly. I tried to consider this topic in such a way as to interest students to a greater extent, to show them the beauty of mathematics from this point of view.

Bibliography:

1. A. Ya. Kannel-Belov, A. K. Kovaldzhi. How to solve non-standard problems Moscow ICSME 2001

2. A.V. Spivak. Supplement to the journal Kvant No. 4/2000 Mathematical holiday, Moscow 2000

3. A.V. Spivak. Mathematical circle, “Sowing” 2003

4. St. Petersburg city ​​palace of youth creativity. Mathematical circle. Problem book for the first and second year of study. Saint Petersburg. 1993

5. Algebra for 8th grade. A textbook for students in schools and classes with in-depth study of mathematics. Edited by N.Ya. Vilenkin. Moscow, 1995

6. M.L. Galitsky, A.M. Goldman, L.I. Zvavich. Collection of algebra problems for 8-9 grades. A textbook for students in schools and classes with in-depth study of mathematics. Moscow, Enlightenment. 1994

7. Yu.N.Makarychev, N.G.Mindyuk, K.I.Neshkov. Algebra 8th grade. A textbook for schools and classes with in-depth study of mathematics. Moscow, 2001

8. M.I.Shabunin, A.A.Prokofiev UMK MATHEMATICS Algebra. Beginnings of mathematical analysis. Profile level. Textbook for 11th grade. Moscow Binom. Knowledge Laboratory 2009

9. M.I. Shabunin, A.A. Prokofiev, T.A. Oleinik, T.V. Sokolova. UMK MATHEMATICS Algebra. Beginnings of mathematical analysis. Profile level Problem book for 11th grade. Moscow Binom. Knowledge Laboratory 2009

10. A.G. Klovo, D.A. Maltsev, L.I. Abzelilova Mathematics. Collection of tests according to the Unified State Exam 2010 plan

11. Unified State Exam-2010. "Legion-M". Rostov-on-Don 2009

12. Unified State Exam UMK “Mathematics. Preparation for the Unified State Exam." Edited by F.F. Lysenko, S.Yu. Kulabukhov. Preparing for Unified State Exam 2011. "Legion-M". Rostov-on-Don 2010

13. UMK "Mathematics. Unified State Exam 2010". Edited by F.F. Lysenko, S.Yu. Kulabukhov. MATHEMATICS Preparation for the Unified State Exam-2010. Educational and training tests. "Legion-M". Rostov-on-Don 2009

14. FIPI Unified State Exam. Universal materials for preparing students MATH 2010"Intellect-Center" 2010

15. A.Zh.Zhafyarov. Mathematics. Unified State Exam-2010 Express consultation. Siberian University Publishing House, 2010

Solving equations in integers.

Uncertain equations are equations containing more than one unknown. By one solution to an indeterminate equation we mean a set of values ​​of the unknowns that turns the given equation into a true equality.

To solve in integers an equation of the form ah + by = c , Where A, b , c - integers other than zero, we present a number of theoretical provisions that will allow us to establish a decision rule. These provisions are also based on already known facts of the theory of divisibility.

Theorem 1.If gcd (A, b ) = d , then there are such integers X And at, that the equality holds ah + b y = d . (This equality is called a linear combination or a linear representation of the greatest common divisor of two numbers in terms of the numbers themselves.)

The proof of the theorem is based on using the equality of the Euclidean algorithm to find the greatest common divisor of two numbers (the greatest common divisor is expressed in terms of partial quotients and remainders, starting from the last equality in the Euclidean algorithm).

Example.

Find the linear representation of the greatest common divisor of the numbers 1232 and 1672.

Solution.

1. Let’s create the equalities of the Euclidean algorithm:

1672 = 1232 ∙1 + 440,

1232 = 440 ∙ 2 + 352,

440 = 352 ∙ 1 + 88,

352 = 88 ∙ 4, i.e. (1672.352) = 88.

2) Let us express 88 sequentially through incomplete quotients and remainders, using the equalities obtained above, starting from the end:

88 = 440 - 352∙1 = (1672 - 1232) - (1232 - 1672∙2 + 1232∙2) = 1672∙3 - 1232∙4, i.e. 88 = 1672∙3 + 1232∙(-4).

Theorem 2. If the equation ah + b y = 1 , if gcd (A, b ) = 1 , it is enough to imagine the number 1 as a linear combination of numbers a and b.

The validity of this theorem follows from Theorem 1. Thus, in order to find a single integer solution to the equation ah + b y = 1, if gcd (a, b) = 1, it is enough to represent the number 1 as a linear combination of numbers A And V .

Example.

Find an integer solution to the equation 15x + 37y = 1.

Solution.

1. 37 = 15 ∙ 2 + 7,

15 = 7 ∙ 2 + 1.

2. 1 = 15 - 7∙2 = 15 - (37 - 15∙2) ∙2 = 15∙5 + 37∙(-2),

Theorem 3. If in Eq. ah + b y = c gcd(a, b ) = d >1 And With not divisible by d , then the equation has no integer solutions.

To prove the theorem, it is enough to assume the opposite.

Example.

Find an integer solution to the equation 16x - 34y = 7.

Solution.

(16.34)=2; 7 is not divisible by 2, the equation has no integer solutions

Theorem 4. If in Eq. ah + b y = c gcd(a, b ) = d >1 and c d , then it is

When proving the theorem, it should be shown that an arbitrary integer solution to the first equation is also a solution to the second equation and vice versa.

Theorem 5. If in Eq. ah + b y = c gcd(a, b ) = 1, then all integer solutions to this equation are contained in the formulas:

t – any integer.

When proving the theorem, it should be shown, firstly, that the above formulas actually provide solutions to this equation and, secondly, that an arbitrary integer solution to this equation is contained in the above formulas.

The above theorems allow us to establish the following rule for solving the equation in integers ah+ b y = c gcd(a, b ) = 1:

1) An integer solution to the equation is found ah + b y = 1 by representing 1 as a linear combination of numbers A Andb (there are other ways to find entire solutions to this equation, for example using continued fractions);

Giving t certain integer values, you can obtain partial solutions to this equation: the smallest in absolute value, the smallest positive (if possible), etc.

Example.

Find integer solutions to the equation 407x - 2816y = 33.

Solution.

1. We simplify this equation, bringing it to the form 37x - 256y = 3.

2. Solve the equation 37x - 256y = 1.

256 = 37∙ 6 + 34,

37 = 34 ∙1 + 3,

34 = 3 ∙11 + 1.

1 = 34 - 3∙11 = 256 - 37∙6 - 11 (37 – 256 + 37∙6) = 256∙12 - 37∙83 =

37∙(-83) - 256∙(-12),

3. General view of all integer solutions of this equation:

x = -83∙3 - 256 t = -249 - 256 t,

y = -12∙3 - 37 t = -36 - 37 t.

The method of exhaustive enumeration of all possible values ​​of variables,

included in the equation.

Find the set of all pairs of natural numbers that are solutions to the equation 49x + 51y = 602.

Solution:

Let us express the variable x from the equation through y x =, since x and y are natural numbers, then x =602 - 51у ≥ 49, 51у≤553, 1≤у≤10.

A complete search of options shows that the natural solutions to the equation are x=5, y=7.

Answer: (5;7).

Solving equations using the factorization method.

Diophantus, along with linear equations, considered quadratic and cubic indefinite equations. Solving them is usually difficult.

Let's consider a case where the difference of squares formula or another method of factorization can be applied to the equations.

Solve the equation in whole numbers: x 2 + 23 = y 2

Solution:

Let's rewrite the equation in the form: y 2 - x 2 = 23, (y - x)(y + x) = 23

Since x and y are integers and 23 is a prime number, the following cases are possible:

Solving the resulting systems, we find:

(-11;12),(11;12),(11;-12),(-11;-12)

Expressing one variable in terms of another and isolating the whole part of the fraction.

Solve the equation in whole numbers: x 2 + xy – y – 2 = 0.

Solution:

Let us express y through x from this equation:

y(x - 1) =2 - x 2,

Nonlinear equations with two unknowns

Definition 1. Let A be some set of pairs of numbers (x; y) . They say that the set A is given numeric function z from two variables x and y , if a rule is specified with the help of which each pair of numbers from set A is associated with a certain number.

Specifying a numerical function z of two variables x and y is often denote So:

Where f (x , y) – any function other than a function

f (x , y) = ax+by+c ,

where a, b, c are given numbers.

Definition 3. Solving equation (2) call a pair of numbers ( x; y) , for which formula (2) is a true equality.

Example 1. Solve the equation

Since the square of any number is non-negative, it follows from formula (4) that the unknowns x and y satisfy the system of equations

the solution to which is a pair of numbers (6; 3).

Answer: (6; 3)

Example 2. Solve the equation

Therefore, the solution to equation (6) is infinite number of pairs of numbers kind

(1 + y ; y) ,

where y is any number.

linear

Definition 4. Solving a system of equations

call a pair of numbers ( x; y) , when substituting them into each of the equations of this system, the correct equality is obtained.

Systems of two equations, one of which is linear, have the form

g(x , y)

Example 4. Solve system of equations

Solution . Let us express the unknown y from the first equation of system (7) through the unknown x and substitute the resulting expression into the second equation of the system:

Solving the equation

x 1 = - 1 , x 2 = 9 .

Hence,

y 1 = 8 - x 1 = 9 ,
y 2 = 8 - x 2 = - 1 .

Systems of two equations, one of which is homogeneous

Systems of two equations, one of which is homogeneous, have the form

where a, b, c are given numbers, and g(x , y) – function of two variables x and y.

Example 6. Solve system of equations

Solution . Let's solve the homogeneous equation

3x 2 + 2xy - y 2 = 0 ,

3x 2 + 17xy + 10y 2 = 0 ,

treating it as a quadratic equation with respect to the unknown x:

.

In case x = - 5y, from the second equation of system (11) we obtain the equation

5y 2 = - 20 ,

which has no roots.

In case

from the second equation of system (11) we obtain the equation

,

whose roots are numbers y 1 = 3 , y 2 = - 3 . Finding for each of these values ​​y the corresponding value x, we obtain two solutions to the system: (- 2 ; 3) , (2 ; - 3) .

Answer: (- 2 ; 3) , (2 ; - 3)

Examples of solving systems of equations of other types

Example 8. Solve a system of equations (MIPT)

Solution . Let us introduce new unknowns u and v, which are expressed through x and y according to the formulas:

In order to rewrite system (12) in terms of new unknowns, we first express the unknowns x and y in terms of u and v. From system (13) it follows that

Let us solve the linear system (14) by eliminating the variable x from the second equation of this system. For this purpose, we perform the following transformations on system (14).

There are many paths leading from the edge of the forest into the thicket. They are tortuous, they converge, diverge again and intersect with each other again. While walking, you can only notice the abundance of these paths, walk along some of them and trace their direction into the depths of the forest. To seriously study the forest, you need to follow the paths until they are visible at all among the dry pine needles and bushes.

Therefore, I wanted to write a project that can be considered as a description of one of the possible walks along the edge of modern mathematics.

The surrounding world, the needs of the national economy, and often everyday worries pose more and more new tasks for a person, the solution to which is not always obvious. Sometimes a particular question has many possible answers, which makes it difficult to solve the problems. How to choose the right and optimal option?

The solution of uncertain equations is directly related to this issue. Such equations, containing two or more variables, for which it is necessary to find all integer or natural solutions, have been considered since ancient times. For example, the Greek mathematician Pythagoras (IV century BC). the Alexandrian mathematician Diophantus (II-III century AD) and the best mathematicians of a closer era to us - P. Fermat (XVII century), L. Euler (XVIII century), J. L. Lagrange (XVIII century) and others.

Participating in the Russian correspondence competition > in Obninsk, the International Game Competition > and the Olympiad of the Ural Federal District, I often encounter such tasks. This is due to the fact that their solution is creative. The problems that arise when solving equations in integers are caused both by complexity and by the fact that little time is devoted to them in school.

Diophantus presents one of the most difficult mysteries in the history of science. We do not know the time when he lived, nor his predecessors who would have worked in the same field. His works are like a sparkling fire in the midst of impenetrable darkness.

The period of time when Diophantus could have lived is half a millennium! The lower bound is determined without difficulty: in his book on polygonal numbers, Diophantus repeatedly mentions the mathematician Hypsicles of Alexandria, who lived in the middle of the 2nd century. BC e.

On the other hand, in the comments of Theon of Alexandria to the famous astronomer Ptolemy there is an excerpt from the work of Diophantus. Theon lived in the middle of the 4th century. n. e. This determines the upper bound of this interval. So, 500 years!

The French historian of science Paul Tannry, editor of the most complete text of Diophantus, tried to narrow this gap. In the Escurial library he found excerpts from a letter from Michael Psellus, a Byzantine scholar of the 11th century. , where it is said that the most learned Anatoly, after collecting the most essential parts of this science, we are talking about the introduction of degrees of the unknown and their (designation), dedicated them to his friend Diophantus. Anatoly of Alexandria actually composed >, excerpts of which are cited in the extant works of Iamblichus and Eusenius. But Anatoly lived in Alexandria in the middle of the 111th century BC. e and even more precisely - until 270, when he became the bishop of Laodacia. This means that his friendship with Diophantus, whom everyone calls Alexandria, must have taken place before this. So, if the famous Alexandrian mathematician and Anatoly's friend named Diophantus are one person, then the time of Diophantus's life is the middle of the 111th century AD.

But the place of residence of Diophantus is well known - Alexandria, the center of scientific thought and the Hellenistic world.

One of the epigrams of the Palatine Anthology has survived to this day:

The ashes of Diophantus rest in the tomb: marvel at it - and the stone

The age of the deceased will speak through his wise art.

By the will of the gods, he lived a sixth of his life as a child.

And I met half past five with fluff on my cheeks.

It was only the seventh day when he became engaged to his girlfriend.

After spending five years with her, the sage waited for his son.

His father's beloved son lived only half his life.

He was taken from his father by his early grave.

Twice for two years the parent mourned severe grief.

Here I saw the limit of my sad life.

Using modern methods of solving equations, it is possible to calculate how many years Diophantus lived.

Let Diophantus live x years. Let's create and solve the equation:

Let's multiply the equation by 84 to get rid of fractions:

Thus, Diophantus lived 84 years.

The most mysterious is the work of Diophantus. Six of the thirteen books that were combined into > have reached us; the style and content of these books differ sharply from the classical ancient works on number theory and algebra, examples of which we know from > Euclid, his >, lemmas from the works of Archimedes and Apollonius. > was undoubtedly the result of numerous studies that remained completely unknown.

We can only guess about its roots, and marvel at the richness and beauty of its methods and results.

> Diophanta is a collection of problems (189 in total), each of which has a solution. The problems in it are carefully selected and serve to illustrate very specific, strictly thought out methods. As was customary in ancient times, methods are not formulated in a general form, but are repeated to solve similar problems.

A unique biography of Diophantus is reliably known, which, according to legend, was carved on his tombstone and represented a puzzle task:

This puzzle serves as an example of the problems that Diophantus solved. He specialized in solving problems in integers. Such problems are currently known as Diophantine problems.

The study of Diophantine equations is usually associated with great difficulties.

In 1900, at the World Congress of Mathematicians in Paris, one of the world's leading mathematicians, David Hilbert, identified 23 problems from various areas of mathematics. One of these problems was the problem of solving Diophantine equations. The problem was the following: is it possible to solve an equation with an arbitrary number of unknowns and integer coefficients in a certain way - using an algorithm. The task is as follows: for a given equation, you need to find all the integer or natural values ​​of the variables included in the equation, at which it turns into a true equality. Diophantus came up with many different solutions for such equations. Due to the infinite variety of Diophantine equations, there is no general algorithm for solving them, and for almost each equation one has to invent an individual technique.

A Diophantine equation of the 1st degree or a linear Diophantine equation with two unknowns is an equation of the form: ax+by=c, where a,b,c are integers, GCD(a,b)=1.

I will give the formulations of theorems on the basis of which an algorithm for solving indeterminate first degree equations of two variables in integers can be compiled.

Theorem 1. If in an equation, then the equation has at least one solution.

Proof:

We can assume that a >0. Having solved the equation for x, we get: x = c-vua. I will prove that if in this formula instead of y we substitute all natural numbers less than a and 0, i.e. the numbers 0;1;2;3;. ;a-1, and each time you perform division, then all a remainders will be different. Indeed, instead of y I will substitute the numbers m1 and m2, smaller than a. As a result, I will get two fractions: c-bm1a and c-bm2a. Having performed the division and denoted the incomplete quotients by q1 and q2, and the remainders by r1 and r2, I will find с-вm1а=q1+ r1а, с-вm2а= q2+ r2а.

I will assume that the remainders r1 and r2 are equal. Then, subtracting the second from the first equality, I get: c-bm1a- c-bm2a = q1-q2, or b(m1 - m2)a = q1-q2.

Since q1-q2 is an integer, then the left side must be an integer. Therefore, bm1 - m2 must be divisible by a, i.e., the difference of two natural numbers, each of which is less than a, must be divisible by a, which is impossible. This means that the remainders r1 and r2 are equal. That is, all residues are different.

That. I received a of various balances less than a. But the distinct a of natural numbers not exceeding a are the numbers 0;1;2;3;. ;a-1. Consequently, among the remainders there will certainly be one and only one equal to zero. The value of y, the substitution of which into the expression (c-vu)a gives a remainder of 0, and turns x=(c-vu)a into an integer. Q.E.D.

Theorem 2. If in the equation, and c is not divisible by, then the equation has no integer solutions.

Proof:

Let d=GCD(a;b), so that a=md, b=nd, where m and n are integers. Then the equation will take the form: mdх+ ndу=с, or d(mх+ nу)=с.

Assuming that there are integers x and y that satisfy the equation, I find that the coefficient c is divisible by d. The resulting contradiction proves the theorem.

Theorem 3. If in the equation, and, then it is equivalent to the equation in which.

Theorem 4. If in an equation, then all integer solutions to this equation are contained in the formulas:

where x0, y0 is an integer solution to the equation, is any integer.

The formulated theorems make it possible to construct the following algorithm for solving an equation of the form in integers.

1. Find the greatest common divisor of the numbers a and b; if c is not divisible by, then the equation has no integer solutions; if and then

2. Divide the equation term by term, obtaining an equation in which.

3. Find an integer solution (x0, y0) of the equation by representing 1 as a linear combination of numbers and;

4. Create a general formula for integer solutions to this equation, where x0, y0 is an integer solution to the equation, and is any integer.

2. 1 DESCENT METHOD

Many > are based on methods for solving uncertain equations. For example, a trick involving guessing the date of birth.

Invite your friend to guess his birthday by the sum of numbers equal to the product of his date of birth by 12 and the number of the month of birth by 31.

In order to guess the birthday of your friend you need to solve the equation: 12x + 31y = A.

Let you be given the number 380, i.e. we have the equation 12x + 31y = 380. In order to find x and y, you can reason like this: the number 12x + 24y is divisible by 12, therefore, according to the properties of divisibility (Theorem 4.4), the number 7y and 380 must have the same remainder when divided by 12. The number 380 when divided by 12 gives a remainder of 8, therefore 7y when divided by 12 must also leave a remainder of 8, and since y is the number of the month, then 1

The equation we solved is a 1st degree Diophantine equation with two unknowns. To solve such equations, the so-called descent method can be used. I will consider the algorithm of this method using the specific equation 5x + 8y = 39.

1. I will choose the unknown that has the smallest coefficient (in our case it is x), and express it through another unknown:. I'll highlight the whole part: Obviously, x will be an integer if the expression turns out to be an integer, which, in turn, will be the case when the number 4 - 3y is divisible by 5 without a remainder.

2. I will introduce an additional integer variable z as follows: 4 - 3y = 5z. As a result, I will get an equation of the same type as the original one, but with smaller coefficients. I will solve it with respect to the variable y:. Selecting the whole part, I get:

Reasoning similarly to the previous one, I introduce a new variable u: 3u = 1 - 2z.

3. I will express the unknown with the smallest coefficient, in this case the variable z: =. Requiring that it be an integer, I get: 1 - u = 2v, whence u = 1 - 2v. There are no more fractions, the descent is complete.

4. Now you need >. I will express through the variable v first z, then y and then x: z = = = 3v - 1; = 3 - 5v.

5. The formulas x = 3+8v and y = 3 - 5v, where v is an arbitrary integer, represent the general solution to the original equation in integers.

Comment. Thus, the descent method involves first sequentially expressing one variable in terms of another until there are no fractions left in the representation of the variable, and then sequentially along a chain of equalities to obtain a general solution to the equation.

2. 2 SURVEY METHOD

Rabbits and pheasants sit in a cage; they have 18 legs in total. Find out how many of both are in the cell?

Let me create an equation with two unknowns, in which x is the number of rabbits, and y is the number of pheasants:

4x + 2y = 18, or 2x + y = 9.

Answer. 1) 1 rabbit and 7 pheasants; 2) 2 rabbits and 5 pheasants; 3) 3 rabbits and 3 pheasants; 4) 4 rabbits and 1 pheasant.

1. PRACTICAL PART

3.1 Solving linear equations with two unknowns

1. Solve the equation 407x - 2816y = 33 in whole numbers.

I will use the compiled algorithm.

1. Using the Euclidean algorithm, I will find the greatest common divisor of the numbers 407 and 2816:

2816 = 407 6 + 374;

407 = 374 1 + 33;

374 = 33 11 + 11;

Therefore (407.2816) = 11, with 33 divisible by 11.

2. Divide both sides of the original equation by 11, we get the equation 37x - 256y = 3, and (37, 256) = 1

3. Using the Euclidean algorithm, I will find a linear representation of the number 1 through the numbers 37 and 256.

256 = 37 6 + 34;

I will express 1 from the last equality, then successively going up the equalities I will express 3; 34 and substitute the resulting expressions into the expression for 1.

1 = 34 - 3 11 = 34 - (37 - 34 1) 11 = 34 12 - 37 11 = (256 - 37 6) 12 - 37 11 =

83 37 - 256 (- 12)

Thus, 37·(- 83) - 256·(- 12) = 1, therefore the pair of numbers x0 = - 83 and y0 = - 12 is a solution to the equation 37x - 256y = 3.

4. I will write down the general formula for solutions to the original equation where t is any integer.

Answer. (-83c+bt; -12c-at), t є Z.

Comment. It can be proven that if the pair (x1,y1) is an integer solution to the equation where, then all integer solutions to this equation are found using the formulas: x=x1+bty=y1-at

2. Solve the equation 14x - 33y=32 in whole numbers.

Solution: x = (32 + 33y) : 14

(14 [. ] 2+ 5)y + (14 [. ] 2 + 4) = 14 [. ] 2y + 5y + 14[. ] 2 + 4 = 14(2y + 2) + 5y + 4; 2y + 2 = p; p є Z

Search from 1 to 13

When y = 2; (5 [. ] 2 + 4): 14

Let me substitute y = 2 into the original equation

14x = 32 +33 [. ] 2

14x = 32 + 66 x = 98: 14 = 7

I will find all the integer solutions from the found quotient:

14(x - 7) + 98 - 33 (y -2) - 66 = 32

14(x - 7) - 33(y - 2)=0

14(x - 7) = 33(y - 2) -> 14(x - 7) : 33 -> (x - 7): 33 -> x = 33k + 7; k є Z

Let me substitute into the original equation:

14(33k + 7) - 33y = 32

14. 33k + 98 - 33y = 32 y = 14k + 2; x = 33k + 7, where k є Z. These formulas specify the general solution to the original equation.

Answer. (33k + 7; 14k + 2), k є Z.

3. Solve the equation x - 3y = 15 in whole numbers.

I will find GCD(1,3)=1

I will determine a particular solution: x=(15+3y):1 using the enumeration method, I find the value y=0 then x=(15+3 [. ] 0) =15

(15; 0) - private solution.

All other solutions are found using the formulas: x=3k + 15, k є Z y=1k+0=k, k є Z for k=0, I get a particular solution (15;0)

Answer: (3k+15; k), k є Z.

4. Solve the equation 7x - y = 3 in whole numbers.

I will find GCD(7, -1)=1

I will define a particular solution: x = (3+y):7

Using the brute force method, we find the value y є y = 4, x = 1

This means (1;4) is a particular solution.

I find all other solutions using the formulas: x = 1k + 1, k є Z y = 7k + 4, k є Z

Answer: (k+1;7k+4); k є Z.

5. Solve the equation 15x+11 y = 14 integers.

I will find GCD(15, -14)=1

I will define a particular solution: x = (14 - 11y):15

Using the brute force method, I find the value y є y = 4, x = -2

(-2;4) is a particular solution.

I find all other solutions using the formulas: x = -11k - 2, k є Z y =15k + 4, k є Z

Answer: (-11k-2; 15k+4); k є Z.

6. Solve the equation 3x - 2y = 12 integers.

I will find GCD(3; 2)=1

I will define a particular solution: x = (12+2y):3

Using the brute force method, I find the value y є y = 0, x = 4

(4;0) is a particular solution.

I find all other solutions using the formulas: x = 2k + 4, k є Z y = 3k, k є Z

Answer: (2k+4; 3k); k є Z.

7. Solve the equation xy = x + y in whole numbers.

I have xy - x - y + 1 = 1 or (x - 1)(y - 1) = 1

Therefore x - 1 = 1, y - 1 = 1, whence x = 2, y = 2 or x - 1 = - 1, y - 1 = - 1, whence x = 0, y = 0 other solutions in integers given the equation does not have.

Answer. 0;0;(2;2).

8. Solve the equation 60x - 77y = 1 in whole numbers.

Let me solve this equation for x: x = (77y + 1) / 60 = (60y + (17y +1)) / 60 = y + (17y + 1) / 60.

Let (17y + 1) / 60 = z, then y = (60z - 1) / 17 = 3z + (9z - 1) / 17. If we denote (9z - 1) / 17 by t, then z = (17t + 1) / 9 = 2t + (- t + 1) / 9. Finally, let (- t + 1) / 9 = n, then t = 1- 9n. Since I find only integer solutions to the equation, z, t, n must be integers.

Thus, z = 2 - 18n + 2 = 2 - 17n, and therefore y = 6 - 51n + 1 - 9n = 7 - 60n, x = 2 - 17n +7 - 60n = 9 - 77n. So, if x and y are integer solutions of a given equation, then there is an integer n such that x = 9 - 77n, y = 7 - 60n. Conversely, if y = 9 - 77n, x = 7 - 60n, then, obviously, x, y are integers. Check shows that they satisfy the original equation.

Answer. (9 - 77n; 7 - 60n)); n є Z.

9. Solve the equation 2x+11y =24 in whole numbers.

I will find GCD(2; 11)=1

I will define a particular solution: x = (24-11y):2

Using the brute force method, I find the value y є y = 0, x = 12

(12;0) is a particular solution.

I find all other solutions using the formulas: x = -11k + 12, k є Z y = 2k + 0=2k, k є Z

Answer:(-11k+12; 2k); k є Z.

10. Solve the equation 19x - 7y = 100 in whole numbers.

I will find GCD(19, -7)=1

I will define a particular solution: x = (100+7y):19

Using the brute force method, I find the value y є y = 2, x = 6

(6;2) is a particular solution.

I find all other solutions using the formulas: x = 7k + 6, k є Z y = 19k + 2, k є Z

Answer:(7k+6; 19k+2); kє Z.

11. Solve the equation 24x - 6y = 144 in whole numbers

I will find GCD(24, 6)=3.

The equation has no solutions because GCD(24, 6)!=1.

Answer. There are no solutions.

12. Solve the equation in whole numbers.

I transform the ratio of coefficients for unknowns.

First of all, I will highlight the whole part of the improper fraction;

I will replace the proper fraction with an equal fraction.

Then I'll get it.

I will do the same transformations with the improper fraction obtained in the denominator.

Now the original fraction will take the form:

Repeating the same reasoning for the fraction, I get.

Isolating the whole part of the improper fraction, I come to the final result:

I got an expression called a finite continued fraction or continued fraction. Having discarded the last link of this continued fraction - one fifth, I will transform the resulting new continued fraction into a simple one and subtract it from the original fraction.

I will reduce the resulting expression to a common denominator and discard it, then

From comparing the resulting equality with the equation it follows that, will be a solution to this equation and, according to the theorem, all its solutions will be contained in,.

Answer. (9+52t; 22+127t), t є Z.

The result obtained suggests that in the general case, in order to find a solution to the equation, it is necessary to expand the ratio of the coefficients of the unknowns into a continued fraction, discard its last link and carry out calculations similar to those carried out above.

13. Solve the equation 3xy + 2x + 3y = 0 in integers.

3xy + 2x + 3y = 3y + 2x + 3y + 2 - 2 = 3y(x + 1) + 2(x + 1) - 2 =

=(x + 1)(3y + 2) - 2,

(x + 1)(3y + 2) = 2,

3y + 2 = 1 or 3y + 1 = 2 or 3y + 1 = -1 or 3y + 1 = -2 x + 1 = 2, x + 1 =1, x + 1 = -2, x + 1 = -1 ; x = 2 or x = 0 or x = -3 or x = -2 y cent z, y = 0, y = -1, y cent z.

Answer: (0;0);(-3;-1).

14. Solve the equation y - x - xy = 2 in whole numbers.

Solution: y - xy - x + 1 = 3, (y + 1)(1 - x) = 3,

3 = 1·3 = 3·1 = (-1)·(-3) = (-3)·(-1).

y + 1 = 1 or y + 1 = 3 or y + 1 = -1 or y + 1 = -3

1 - x =3, 1 - x =1, 1 - x = -3, 1 - x = -1.

y = 0 or y = 2 or y = -2 or y = -4 x = -2, x = 0, x = 4, x = 2

Answer: (-2;0);(0;2);(2;-4);(4;-2).

15. Solve the equation y + 4x + 2xy = 0 in whole numbers.

Solution: y + 4x + 2xy + 2 - 2 = 0, (2x + 1)(2 + y) = 2,

2 = 1∙2 = 2∙1 = (-2)∙(-1) = (-1)∙(-2).

2x + 1= 1 or 2x + 1= 2 or 2x + 1= -1 or 2x + 1= -2

2 + y = 2, 2 + y = 1, 2 + y = -2, 2 + y = -1; y = 0 or y = -1 or y = -4 or y = -3 x = 0, x cent Z, x = -1, x cent Z.

Answer: (-1;-4);(0;0).

16. Solve the equation 5x + 10y = 21 in whole numbers.

5(x + 2y) = 21, since 21 != 5n, then there are no roots.

Answer. There are no roots.

17. Solve the equation 3x + 9y = 51 in natural numbers.

3(x + 3y) = 3∙17, x = 17 - 3y, y = 1, x = 14; y = 2, x = 11; y = 3, x = 8; y = 4, x = 5; y = 5, x = 2; y = 6, x = -1, -1cent N.

Answer:(2;5);(5;4);(8;3);(11;2; (14:1).

18. Solve the equation 7x+5y=232 in whole numbers.

I will solve this equation with respect to the unknown at which the smallest (modulo) coefficient is found, that is, in this case with respect to y: y = 232-7x5.

Let me substitute the numbers instead of x into this expression: 0;1;2;3;4. I get: x=0, y=2325=4625, x=1, y=232-75=45, x=2, y=232-145=43.6, x=3, y=232-215=42, 2, x=4, y=232-285=40.8

Answer. (1;45).

19. Solve the equation 3x + 4y + 5xy = 6 in whole numbers.

I have 3∙4 + 5∙6 = 42 = mn

Divisors 42: - +- (1, 2, 3, 6, 7, 14, 21, 42).

x = m - 45, y = n - 35 I find that for m = -1, -6, 14, -21 n = -42, -7, 3, -2 the solutions are: x = -1, -2, 2, -5 y = -9, -2, 0, -1.

So, this equation has 4 solutions in integers and none in natural numbers.

Answer. -1;-9;-2;-2;2;0;(-5;-1).

20. Solve the equation 8x+65y=81 in natural numbers.

81⋮GCD(8;65)=>

8x=81-65y x=81-65y8=16+65-65y8=2+65(1-y)8.

Let 1-y8=t, t Є Z. x=2+65t>0y=1-8t>0

65t>-2-8t>-1 t>-265 t t=0.

At t=0 x=2y=1

Answer. (2;1).

21. Find integer non-negative solutions to the equation 3x+7y=250.

250⋮GCD(3;7) =>the equation can be solved in integers.

x=250-7y3=243+7-7y3=81+7(1-y)3.

Let 1-y3=t, t Є Z.

x=81+7t>=0y=1-3t>=0

7t>=-81-3t>=-1 t>=-817t=-1147t t=-11;-10;. ;0.

x=81+7tу=1-3t t=-11 ​​x=4y=34 t=-10 x=11y=31 t=-9 x=18y=28 t=-8 x=25y=25 t=-7 x =32y=22 t=-6 x=39y=19 t=-5 x=46y=16 t=-4 x=53y=13 t=-3 x=60y=10 t=-2 x=67y=7 t =-1 x=74y=4 t=0 x=81y=1

Answer. 11;31;18;28;25;25;32;22;39;19;46;16;53;13;60;10;67;7;74;4;81;1.

22. Solve the equation xy+x+y3=1988 in integers.

Let's multiply both sides of the equation by 3. We get:

3x+3xy+y=5964

3x+3xy+y+1=5965

(3х+1)+(3х+у)=5965

(3x+1) + y(3x+1)=5965

(3x+1)(y+1)=5965

5965=1∙5965 or 5965=5965∙1 or 5965=-1∙(-5965) or 5965=-5965∙(-1) or 5965=5∙1193 or 5965=1193∙1 or 5965=-5∙( -1193) or 5965=-1193∙(-5)

1) 3x+1=1y+1=5965 2) 3x+1=5965y+1=1 x=0y=5964 x=1988y=0

3) 3x+1=5y+1=1193 4) 3x+1=1193y+1=5 solutions in integers no solutions in integers no

5) 3x+1=-1y+1=-5965 6) 3x+1=-5965y+1=-1 no solutions in integers no solutions in integers

7) 3x+1=-5y+1=-1193 8) 3x+1=-1193y+1=-5 x=-2y=1194 x=-398y=-6

Answer. 0;5964;1988;0;-2;-1194;(-398;-6).

3. 2 SOLVING PROBLEMS

There are several types of problems, most often these are problems of an Olympiad nature, which boil down to solving Diophantine equations. For example: a) Tasks on exchanging a sum of money of a certain denomination.

b) Problems involving transfusion and dividing objects.

1. We bought 390 colored pencils in boxes of 7 and 12 pencils. How many of these and other boxes did you buy?

I will designate: x boxes of 7 pencils, y boxes of 12 pencils.

Let me create an equation: 7x + 12y = 390

I will find GCD(7, 12)=1

I will define a particular solution: x = (390 - 12y):7

Using the brute force method, I find the value y є y = 1, x = 54

(54;1) is a particular solution.

I find all other solutions using the formulas: x = -12k + 54, k є Z y = 7k + 1, k є Z

I found many solutions to the equation. Taking into account the conditions of the problem, I will determine the possible number of both boxes.

Answer. You can buy: 54 boxes of 7 pencils and 1 box of 12 pencils, or 42 boxes of 7 pencils and 8 boxes of 12 pencils, or 30 boxes of 7 pencils and 15 boxes of 12 pencils, or 28 boxes of 7 pencils and 22 boxes of 12 pencils , or 6 boxes of 7 pencils and 29 boxes of 12 pencils.

2. One leg of a right triangle is 7 cm larger than the other, and the perimeter of the triangle is 30 cm. Find all the sides of the triangle.

I will designate: x cm - one leg, (x+7) cm - the other leg, y cm - hypotenuse

I will compose and solve the Diophantine equation: x+(x+7)+y=30

I will find GCD(2; 1)=1

I will define a particular solution: x = (23 - y):2

Using the brute force method, I find the value y =1 y = 1, x = 11

(11;1) is a particular solution.

I find all other solutions to the equation using the formulas: x = -k + 11, k є Z y = 2k + 1, k є Z k

Considering that any side of a triangle is less than the sum of the other two sides, we come to the conclusion that there are three triangles with sides 7, 9 and 14; 6, 11 and 13; 5, 13 and 12. According to the conditions of the problem, a right triangle is given. This is a triangle with sides 5, 13 and 12 (the Pythagorean theorem holds).

Answer: One leg is 5 cm, the other is 12 cm, the hypotenuse is 13 cm.

3. Several children were picking apples. Each boy collected 21 kg, and the girl collected 15 kg. In total they collected 174 kg. How many boys and how many girls picked apples?

Let there be x boys and y girls, with x and y being natural numbers. Let me create an equation:

I solve by selection method: x

6 Only at x = 4 does the second unknown receive a positive integer value (y = 6). For any other value of x, y will be either a fraction or negative. Therefore, the problem has one unique solution.

Answer. 4 boys and 6 girls.

4. Is it possible to create a set of pencils worth 3 rubles and pens worth 6 rubles worth 20 rubles?

Let the number of pencils in the set be x and the number of pens be y.

Let me create an equation:

For any integers x and y, the left side of the equation must be divisible by 3; the right-hand side is not divisible by 3. This means that there are no integers x and y that would satisfy our equation. This equation cannot be solved in integers. It is impossible to create such a set.

Answer. There are no solutions.

5. Find a natural number that, when divided by 3, leaves a remainder of 2, and when divided by 5, leaves a remainder of 3.

I will denote the required number by x. If I denote the quotient of x by 3 by y, and the quotient of division by 5 by z, then I get: x=3y+2x=5z+3

According to the meaning of the problem, x, y and z must be natural numbers. This means that we need to solve an indefinite system of equations in integers.

For any integer y and z, x will also be an integer. I subtract the first from the second equation and get:

5z - 3y + 1 = 0.

Having found all positive integers y and z, I will immediately obtain all positive integer values ​​of x.

From this equation I find:

One solution is obvious: for z = 1 we get y = 2, and x and y are integers. The solution x = 8 corresponds to them.

I'll find other solutions. To do this, I will introduce an auxiliary unknown u, setting z = 1 + u. I will receive:

5(1 + u) - 3y + 1 = 0, i.e. 5u = 3y - 6 or 5u = 3(y - 2).

The right-hand side of the last equation is divisible by 3 for any integer y. This means that the left-hand side must also be divisible by 3. But the number 5 is coprime to the number 3; therefore u must be divisible by 3, i.e., have the form 3n, where n is an integer. In this case, y will equal

15n/3 + 2 = 5n + 2, i.e., also an integer. So, z = 1 + u = 1 + 3n, whence x = 5z + 3 = 8 + 15n.

The result is not one, but an infinite set of values ​​for x: x = 8 + 15n, where n is an integer (positive or zero):

Answer. x=8+15n; n є 0;1;2;.

6. Subjects brought 300 precious stones as a gift to the Shah: in small boxes of 15 pieces each and in large ones - 40 pieces. How many of these and other boxes were there, if it is known that there were fewer small ones than large ones?

Let me denote by x the number of small boxes, and by y the number of large ones.

15x+40y=300. I'll cut it by 5.

3x+8y=60 x=60-8y3 x=60-6y-2y3

X=20-2y-2y3

For the value of a fraction to be an integer, 2y must be a multiple of 3, i.e. 2y = 3c.

I will express the variable y and select the whole part:

Z must be a multiple of 2, i.e. z=2u.

Let me express the variables x and y in terms of u:

X=20-2y-2y3

Х=20-2∙3u-2∙3u3

I will compose and solve a system of inequalities:

I will write down the entire solutions: 1; 2. Now I will find the values ​​of x and y for u=1; 2.

1) x1=20-8∙1=20-8=12 y1=3∙1=3

2) x2=20-8∙2=20-16=4 y2=3∙2=6

Answer. 4 small boxes; 6 large boxes.

7. Two Ural 5557 cars were given, the cars were sent on a flight Krasnoturinsk - Perm - Krasnoturinsk. In total, 4 tons of diesel fuel and 2 drivers were needed to complete this flight. It is necessary to determine transport costs, namely the cost of 1 ton of diesel fuel and wages for drivers performing this flight, if it is known that a total of 76,000 rubles were spent.

Let x rubles be the cost of 1 ton of diesel fuel, and let x rubles be the wages of drivers. Then (4x + 2y) rubles were spent on the flight. And according to the conditions of the problem, 76,000 rubles were spent.

I get the equation:

To solve this equation, the brute-force method will be a labor-intensive process. So I'll use the > method.

I’ll express the variable y through x: , select the whole part, and get: (1).

For the value of a fraction to be an integer, 2x must be a multiple of 4. That is, 2x = 4z, where z is an integer. From here:

I’ll substitute the value of x into expression (1):

Since x, y 0, then 19000 z 0, therefore, giving z integer values ​​from 0 to 19000, I get the following values ​​of x and y: z

From real data on transportation costs, it is known that 1 ton of diesel fuel (x) costs 18,000 rubles. , and the payment for drivers performing flight (y) is 10,000 rubles. (data taken approximately). From the table we find that the x value equal to 18000 and the y value equal to 10000 correspond to a z value equal to 9000, indeed: ;.

8. In how many ways can you collect the amount of 27 rubles? , having quite a lot of two-ruble and five-ruble coins?

Let me denote: x two-ruble coins and y five-ruble coins

I will create an equation, taking into account the condition of the problem 2x + 5y = 27.

I will find GCD(2;5)=1

I will define a particular solution: x = (27-5y):2

Using the brute force method, I find the value y є y = 1, x = 11

(11;1) is a particular solution.

All other solutions are found using the formulas: x = -5k + 11, k є Z y = 2k + 1, k є Z

This equation has many solutions. Let's find all the ways in which you can collect the amount of 27 rubles with the offered coins. k

Answer. There are three ways in which you can collect this amount if you have a lot of two-ruble and five-ruble coins.

9. Let's say octopuses and starfish live in an aquarium. Octopuses have 8 legs, and starfish have 5. There are 39 limbs in total. How many animals are there in the aquarium?

Let x be the number of starfish, y the number of octopuses. Then all octopuses have 8 legs, and all stars have 5 legs.

Let me create an equation: 5x + 8y = 39.

Please note that the number of animals cannot be expressed as non-integer or negative numbers. Therefore, if x is a non-negative integer, then y = (39 - 5x)/8 must also be an integer and non-negative, and, therefore, it is necessary that the expression 39 - 5x be divisible by 8 without a remainder. A simple search of options shows that this possible only when x = 3, then y = 3.

Answer: (3; 3).

10. A furniture factory produces stools with three and four legs. The master made 18 legs. How many stools can be made so that all the legs can be used?

Let x be the number of three-legged stools and y the number of four-legged ones. Then, 3x + 4y = 18.

I have, 4y =18 - 3x; y = 3(6 - x):4.

I get: x = 2; y = 3 or x = 6; y = 0.

There are no other solutions, since x 6.

Answer. 2;3;(6;0).

11. Is it possible to accommodate 718 people in 4- and 8-berth cabins, so that there are no empty seats in the cabins?

Let the 4-bed cabins be x, and the 8-bed cabins y, then:

2(x + 2y) = 309

Answer. It is forbidden.

12. Prove that on the line 124x + 216y = 515 there is not a single point with integer coordinates.

GCD(124,216) = 4, 515 != 4n, which means there are no integer solutions.

Answer. There are no solutions.

13. The cost of the goods is 23 rubles, the buyer has only 2 ruble coins, and the cashier has 5 ruble coins. Is it possible to make a purchase without first exchanging money?

Let x be the number of 2 ruble coins, y the number of 5 ruble coins, then 2x - 5y = 23, where x,y є N.

I get: 2x = 23 + 5y, from where x =23 + 5y2 =11 + 2y + (1 + y)2 x will be an integer if 1 + y2 is an integer.

1 + y2 = t, where t Euro Z, then y = 2t - 1.

x = 11 + 2y + 1 + y2 = 11 + 4t - 2 + 1 + 2t-12 = 5t + 9.

T. o. x = 5t + 9, and y = 2t - 1, where t є z.

The problem has many integer solutions. The simplest of them is for t = 1, x =14, y = 1, i.e. the buyer will give fourteen 2-ruble coins and receive one 5-ruble coin in change.

Answer. Can.

14. During an audit of the store’s trade books, one of the entries turned out to be covered in ink and looked like this:

> It was impossible to make out the number of meters sold, but there was no doubt that the number was not a fraction; in the proceeds it was possible to distinguish only the last three digits, and it was also possible to establish that there were three other digits in front of them. Is it possible to restore a record using this data?

Let the number of meters be x, then the cost of the goods in kopecks is 4936x. We denote the total of three filled in digits as y, this is the number of thousands of kopecks, and the entire amount in kopecks will be expressed as follows (1000y + 728).

I get the equation 4936x = 1000y + 728, I divide it by 8.

617x - 125y = 91, where x,y є z, x,y

125y = 617x - 91 y = 5x - 1 +34 - 8x125 = 5x - 1 + 2 17 - 4x125 =

5x - 1 + 2t, where t = 17 - 4x125, t Euro Z.

From the equation t = (17 - 4x)/125 I get x = 4 - 31t + 1 - t4 =

4 - 31t + t1, where t1 = 1 - t4, hence t = 1 - 4t1, a x = 125t1 - 27, y = 617t1 - 134.

By condition I know that 100

100 = 234/617 and t1

This means that 98 meters were sold for the amount of 4837.28 rubles. The recording has been restored.

Answer. 98 meters released.

15. It is required to buy 40 postage stamps for one ruble - kopeck, 4-kopeck and 12-kopeck. How many stamps of each denomination can you buy?

You can make two equations: x + 4y + 12z = 100 and x + y + z = 40, where x is the number of penny marks, y is the number of 4-kopeck marks, z is the number of 12-kopeck marks. I subtract the second from the first equation and get:

3y + 11z = 60, y = 60 - 11z3 = 20 - 11· z3.

Let z3 = t, z = 3t, where t Euro Z. Then I get if x + y + z = 40 and z = 3t, and y = 20 - 11t, x = 20 + 8t.

Since x >= 0, y >= 0, z >= 0, then 0

Then, accordingly, I get: t = 0, x = 20, y = 20, z = 0; t = 1, x = 28, y = 9, z = 3.

So, the purchase of stamps can be made in only two ways, and if the condition is that at least one stamp of each denomination be purchased, then only in one way.

Answer. 28 marks of 1 kopeck, 9 marks of 4 kopecks and 3 marks of 12 kopecks.

16. A student was given a task of 20 problems. For each correctly solved question, he receives 8 points; for each unsolved question, 5 points are deducted from him. For a task that he did not undertake - 0 points. The student scored 13 points in total. How many problems did he undertake to solve?

Let the correctly solved problems be x, the incorrectly solved problems be y, and the not considered problems be z.

Then x + y + z = 20, and 8x - 5y = 13.

y = 8x - 135= x - 2 +3(x - 1)5 = x - 2 + 3t, ​​where t = x - 15, and x = 5t + 1.

By condition x + y

Answer: the student took on 13 problems, solved 6, and failed 7.

17. Ivanushka the Fool fights with the Serpent Gorynych, who has 2001 heads. Swinging his sword to the left, Ivan cuts off 10 heads, and in return 16 grow. Swinging his sword to the right, he cuts off 15, and 6 grow. If all the heads are cut off, no new ones grow. You can swing in any order, but if there are less than 15 goals, then only to the left, and if there are less than 10, then not at all. Can Ivanushka the Fool defeat the Serpent Gorynych?

Let me rephrase the problem: is it possible to cut down 1986 heads? Then Ivan will cut down the remaining 15 with one blow to the right and no new ones will grow.

Let x be the number of strokes to the right, and y the number of strokes to the left, then 1986 - 9x + 6y = 0.

I divide the whole equation by 6, I get

3x - 2y = 662.

y = 3x - 6622 = x - 331 + x2.

Let x2 = t, then x = 2t, and y = 3t - 331.

Since x >= 0, y >= 0, then t >= 111, hence t = 111, x = 222, y = 2.

I get: by hitting 220 times to the right, Ivan cuts off 1980 heads and the Serpent has 21 heads left; then 2 hits to the left and the Snake grows 12 heads, making a total of 33; the next 2 blows to the right deprive the Snake of 18 heads and Ivan cuts off the remaining 15 with the last blow to the right and no new heads grow.

Answer: 220 strikes to the right, 2 strikes to the left and 3 more strikes to the right.

18. The sides of a dice are numbered - 1, 2, 3, 4, 5, 6. From 5 such cubes, they built a tower and counted the sum of points on all visible faces, after removing the top cube, the sum decreased by 19, which number turned out to be the top edge of the top cube?

The sum of points of one cube is 21.

Let x be the number of points on the bottom edge of the top cube, and y the number of points on the top edge of the next cube. When you remove the top cube, the points of 5 faces of the top cube disappear, the sum of the points of which is (21 - x), and the face on which the points appears, which means that the sum of the points has decreased by (21 - x) - y, and according to the condition it is 19, hence :

(21 - x) - y = 19, x + y = 2.

Hence y = 2 - x, and by condition 1

19. Someone bought 30 birds for 30 coins of the same denomination. For every 3 sparrows you pay 1 coin, for 2 bullfinches - 1 coin, for 1 dove - 2 coins. How many birds of each type were there?

Let there be x sparrows, y bullfinches, and z pigeons. Then, according to the condition, x + y + z = 30 and 13x + 12y + 2z = 30.

I get x + y + z = 30 and 2x + 3y + 12z = 180, or y + 10z = 120, y = 120 - 10z, where by condition x

Hence the following options (0;20;10); (9;10;11); (18;0;12).

Answer: sparrows - 0, bullfinches - 20, pigeons - 10; sparrows - 9, bullfinches - 10, pigeons - 11; sparrows - 18, bullfinches - 0, pigeons - 12.

20. Find all two-digit numbers, each of which, when reduced by 2, is equal to five times the product of its digits.

Let xy be the required two-digit numbers.

For the equation xy - 2 = 5xy, or (10x + y) - 5xy = 2 S = 0 and I will find all natural solutions from the set (x; 2).

Since x is the first digit of two-digit numbers, it can only take 9 values.

That. , the required numbers will be: 12, 22, 32,. , 92.

Answer. 12; 22, 32; 42; 52; 62; 72; 82; 92.

21. A piece of wire 102 cm long needs to be cut into pieces 15 cm and 12 cm long so that all the wire is used. How to do it?

Let x be the number of parts of a wire 15 cm long, y the number of parts of a wire 12 cm long. Let me create an equation:

15x+12y=102 /:3

4x+3y=34 x=34-4y5=6+4-4y5=6+4(1-y)5.

Let 1-y5=t x=6+4t>0y=1-5t>0=> 4t>-6-5t>-1 => t>-1.5t t=0;-1.

If t=0, then x=6y=1

If t=-1, then x=2y=6

Answer. The problem has two solutions:

1) 102=15∙6+12∙1; 2) 102=15∙2+12∙6.

22. Petya in 1987 was as old as the sum of the digits of the year of his birth. In which year he was born?

Let Petya be born in 1919. Then in 1987 he was 1987-19xy, or (1+9+x+y) years old. We have the equation:

87-(10x+y)=10+x+y

77-11x=2y y=77-11x2=38-11x-12.

Considering that x and y are digits of the decimal number system, we find by selection: x=3, y=1.

Answer. Petya was born in 1970.

23. Someone buys an item worth 19 rubles in a store. He has only 15-three-ruble notes, while the cashier has only 20-five-ruble notes. Can I pay and how?

The problem comes down to solving the Diophantine equation in positive integers: 3x - 5y = 19, where x

Due to the fact that x>0 and y > 0 and taking into account the conditions of the problem, it is easy to establish that 0

This leads to 2 possible values: x

Answer. 1) 19=3∙8-1∙5 2) 19=3∙13-4∙5.

24. Is it possible to weigh 28 g of a certain substance on a cup scale, having only 4 weights weighing 3 g and 7 weights weighing 5 g?

To do this you need to solve the equation:

x = 9 - 2(3y1 - 1) + y1 = 11-5y1.

So x = 11 - 5 y1 y = 3 y1 - 1.

It follows from the conditions of the problem that y1 cannot be given negative values. Next should be y1

Answer. 1 weight in 3 g and 5 weights in 5 g.

25. The buyer purchased in the store for 21 rubles. goods. But he only has banknotes of 5 ruble denominations, while the cashier has 3 ruble ones. You want to know if you can pay the cashier if you have money and how exactly?

Let x be the number 5 - rubles, y - 3 - rubles.

By condition, x > 0, y > 0, that means.

Also, t is even, otherwise neither x nor y will be integers.

At t = 4, 6, 8,. we have: t

Answer. 6;3;8;8;12;13;15;18;18;23;21;28;24;33;27;38;(30;43).

26. There are 110 sheets of paper. It is required to sew notebooks of 8 sheets and 10 sheets each. How many do you need to sew?

Let x be the number of 8 sheet notebooks, y the number 10 sheet notebooks.

So t = 0 or t = - 1

Answer. 5;7;(10;3).

27. Many ancient methods of guessing numbers and dates of birth are based on solving Diophantine equations. For example, to guess the date of birth (month and day) of your interlocutor, it is enough to ask him for the sum obtained from adding two products: the date number (x) by 12 and the month number (y) by 31.

Let the sum of the products in question be equal to 330. Find the date of birth.

Let's solve the indeterminate equation: y = 2y1 + y2 = 2(2y2 + y3) + y2 = 5y2 + 2y3 = 5(2y3 - 6) + 2y3 = 12y3 - 30 x = 27 - 3(12y3 - 30) + 2y2 + y3 = 27 - 36y3 + 90 + 2(2y3 - 6) + y3 =

27 - 36y3 + 90 + 5y3 - 12 = 105 - 31y3 x = 12y3 - 30, y = 105 - 31y3

So, date of birth: 12th day of the 6th month.

28. Is it possible to collect the amount of 51 rubles with two-ruble and five-ruble coins? If possible, how many ways are there?

Let there be x two-ruble coins, and five-ruble coins.

Let 1+y2=z, then

=> z = 1, 2, 3, 4, 5

Answer: 5 ways.

29. Is it possible to put two hundred eggs in boxes of 10 and 12 pieces? If possible, find all such ways.

Let there be x boxes of 10 pieces each and let the boxes have 12 pieces each. Let me create an equation: z = 1, 2, 3

Answer: 14;5;8;10;(2;15)

30. Imagine the number 257 as the sum of two natural terms: a) one of which is a multiple of 3, and the other is a multiple of 4; b) one of which is a multiple of 5, and the other is a multiple of 8.

Answer: 1) 249 and 8; 2) 225 and 32.

In problems involving indefinite equations, I encountered a wide variety of cases: the problem may be completely unsolvable (Problem 4), may have an infinite number of solutions (Problem 2), may have several definite solutions; in particular, it can have one unique solution (Problem 1).

CONCLUSION

The goal that I set for myself has been achieved. Working on the project aroused interest and captivated me. This work required from me not only certain mathematical knowledge and perseverance, but also gave me the opportunity to feel the great joy of independent discovery.

Diophantine equations are found in Olympiad tasks, so they develop logical thinking, increase the level of mathematical culture, and instill skills in independent research work in mathematics.

When solving equations and problems that reduce to Diophantine equations, the properties of prime numbers, the method of factoring a polynomial, the enumeration method, the descent method and the Euclidean algorithm are used. In my opinion, the descent method is the most difficult. But the brute force method turned out to be prettier for me.

I solved 54 problems in my work.

This work contributed to a deeper understanding of the school curriculum and broadened my horizons.

This material will be useful to students interested in mathematics. It can be used in some lessons and extracurricular activities.