This article begins with material divisibility theory of integers. Here we introduce the concept of divisibility and indicate the accepted terms and notations. This will allow us to list and justify the main properties of divisibility.
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The concept of divisibility
The concept of divisibility is one of the basic concepts of arithmetic and number theory. We will talk about divisibility and in special cases - about divisibility. So, let's give an idea of divisibility on the set of integers.
Integer a shares by an integer b, which is different from zero, if there is an integer (denote it by q) such that the equality a=b·q is true. In this case we also say that b divides a. In this case, the integer b is called divider numbers a, the integer a is called multiples the number b (for more information on divisors and multiples, see the article Divisors and Multiples), and the integer q is called private.
If an integer a is divisible by an integer b in the above sense, then a can be said to be divisible by b completely. The word “entirely” in this case further emphasizes that the quotient of dividing the integer a by the integer b is an integer.
In some cases, for given integers a and b, there is no integer q for which the equality a=b·q is true. In such cases, we say that the integer a is not divisible by the integer b (meaning that a is not divisible by b). However, in these cases they resort to.
Let's understand the concept of divisibility using examples.
Any integer a is divisible by the number a, by the number −a, a, by one and by the number −1.
Let us prove this property of divisibility.
For any integer a, the equalities a=a·1 and a=1·a are valid, from which it follows that a is divisible by a, and the quotient is equal to one, and that a is divisible by 1, and the quotient is equal to a. For any integer a, the equalities a=(−a)·(−1) and a=(−1)·(−a) are also valid, from which it follows that a is divisible by the number opposite to a, as well as a is divisible by minus unit.
Note that the property of divisibility of an integer a by itself is called the property of reflexivity.
The next property of divisibility states that zero is divisible by any integer b.
Indeed, since 0=b·0 for any integer b, then zero is divisible by any integer.
In particular, zero is also divisible by zero. This confirms the equality 0=0·q, where q is any integer. From this equality it follows that the quotient of zero divided by zero is any integer.
It should also be noted that no other integer a other than zero is divisible by 0. Let's explain this. If zero divided an integer a different from zero, then the equality a=0·q should be true, where q is some integer, and the last equality is possible only if a=0.
If an integer a is divisible by an integer b and a is less than the modulus of b, then a is equal to zero. In literal form, this property of divisibility is written as follows: if ab and , then a=0.
Proof.
Since a is divisible by b, there is an integer q for which the equality a=b·q is true. Then equality must also be true, and by virtue equality of the form must also be true. If q is not equal to zero, then it follows that . Taking into account the obtained inequality, it follows from the equality that . But this contradicts the condition. Thus, q can only be equal to zero, and we obtain a=b·q=b·0=0, which is what we needed to prove.
If an integer a is non-zero and divisible by an integer b, then the modulus of a is not less than the modulus of b. That is, if a≠0 and ab, then . This property of divisibility follows directly from the previous one.
The only divisors of unity are the integers 1 and −1.
First, let's show that 1 is divisible by 1 and −1. This follows from the equalities 1=1·1 and 1=(−1)·(−1) .
It remains to prove that no other integer is a divisor of unity.
Suppose that an integer b, different from 1 and −1, is a divisor of unity. Since unity is divisible by b, then, due to the previous property of divisibility, the inequality must be satisfied, which is equivalent to the inequality. This inequality is satisfied by only three integers: 1, 0, and −1. Since we assumed that b is different from 1 and −1, then only b=0 remains. But b=0 cannot be a divisor of unity (as we showed when describing the second property of divisibility). This proves that no numbers other than 1 and −1 are divisors of unity.
For an integer a to be divisible by an integer b it is necessary and sufficient that the modulus of the number a is divisible by the modulus of the number b.
Let us first prove the necessity.
Let a be divided by b, then there is an integer q such that a=b·q. Then 
. Since it is an integer, the equality implies that the modulus of the number a is divisible by the modulus of the number b.
Now sufficiency.
Let the modulus of the number a be divided by the modulus of the number b, then there exists an integer q such that . If the numbers a and b are positive, then the equality a=b·q is true, which proves the divisibility of a by b. If a and b are negative, then the equality −a=(−b)·q is true, which can be rewritten as a=b·q. If a – a negative number, and b is positive, then we have −a=b·q, this equality is equivalent to the equality a=b·(−q) . If a is positive and b is negative, then we have a=(−b)·q , and a=b·(−q) . Since both q and −q are integers, the resulting equalities prove that a is divisible by b.
Corollary 1.
If an integer a is divisible by an integer b, then a is also divisible by the opposite number −b.
Corollary 2.
If an integer a is divisible by an integer b, then −a is also divisible by b.
The importance of the just discussed property of divisibility is difficult to overestimate - the theory of divisibility can be described on the set of positive integers, and this property of divisibility extends it to negative integers.
Divisibility has the property of transitivity: if an integer a is divisible by some integer m, and the number m in turn is divided by some integer b, then a is divisible by b. That is, if am and mb, then ab.
Let us give a proof of this divisibility property.
Since a is divisible by m, there is some integer a 1 such that a=m·a 1. Similarly, since m is divisible by b, there is some integer m 1 such that m=b·m 1. Then a=m a 1 =(b m 1) a 1 =b (m 1 a 1). Since the product of two integers is an integer, then m 1 ·a 1 is some integer. Denoting it q, we arrive at the equality a=b·q, which proves the property of divisibility under consideration.
Divisibility has the property of antisymmetry, that is, if a is divided by b and at the same time b is divided by a, then either the integers a and b, or the numbers a and −b, are equal.
From the divisibility of a by b and b by a, we can talk about the existence of integers q 1 and q 2 such that a=b·q 1 and b=a·q 2. Substituting b·q 1 instead of a into the second equality, or substituting a·q 2 instead of b into the first equality, we obtain that q 1 ·q 2 =1, and given that q 1 and q 2 are integers, this is only possible if q 1 =q 2 =1 or when q 1 =q 2 =−1. It follows that a=b or a=−b (or, what is the same, b=a or b=−a ).
For any integer and non-zero number b, there is an integer a, not equal to b, that is divisible by b.
This number will be any of the numbers a=b·q, where q is any integer not equal to one. We can move on to the next property of divisibility.
If each of two integer terms a and b is divisible by an integer c, then the sum a+b is also divisible by c.
Since a and b are divisible by c, we can write a=c·q 1 and b=c·q 2. Then a+b=c q 1 +c q 2 =c (q 1 +q 2)(the last transition is possible due to ). Since the sum of two integers is an integer, the equality a+b=c·(q 1 +q 2) proves the divisibility of the sum a+b by c.
This property can be extended to the sum of three, four or more terms.
If we also remember that subtracting an integer b from an integer a is the addition of the number a with the number −b (see), then this property of divisibility is also true for the difference of numbers. For example, if the integers a and b are divisible by c, then the difference a−b is also divisible by c.
If it is known that in an equality of the form k+l+…+n=p+q+…+s all terms except one are divisible by some integer b, then this one term is also divisible by b.
Let's say this term is p (we can take any of the terms of the equality, which will not affect the reasoning). Then p=k+l+…+n−q−…−s . The expression obtained on the right side of the equality is divided by b due to the previous property. Therefore, the number p is also divisible by b.
If an integer a is divisible by an integer b, then the product a·k, where k is an arbitrary integer, is divided by b.
Since a is divisible by b, the equality a=b·q is true, where q is some integer. Then a·k=(b·q)·k=b·(q·k) (the last transition was carried out due to ). Since the product of two integers is an integer, the equality a·k=b·(q·k) proves the divisibility of the product a·k by b.
Corollary: if an integer a is divisible by an integer b, then the product a·k 1 ·k 2 ·…·k n, where k 1, k 2, …, k n are some integers, is divisible by b.
If integers a and b are divisible by c, then the sum of the products a·u and b·v of the form a·u+b·v, where u and v are arbitrary integers, is divided by c.
The proof of this divisibility property is similar to the previous two. From the condition we have a=c·q 1 and b=c·q 2. Then a u+b v=(c q 1) u+(c q 2) v=c (q 1 u+q 2 v). Since the sum q 1 ·u+q 2 ·v is an integer, then an equality of the form a u+b v=c (q 1 u+q 2 v) proves that a·u+b·v is divisible by c.
This concludes our review of the basic properties of divisibility.
Bibliography.
- Vilenkin N.Ya. and others. Mathematics. 6th grade: textbook for general education institutions.
- Vinogradov I.M. Fundamentals of number theory.
- Mikhelovich Sh.H. Number theory.
- Kulikov L.Ya. and others. Collection of problems in algebra and number theory: Tutorial for students of physics and mathematics. specialties of pedagogical institutes.
Name the numbers used for counting. Each number of countable items corresponds to a certain natural number. If there are no objects to count, then the number 0 is used, but when counting objects we never start from 0, and accordingly the number 0 cannot be classified as natural. It is clear that the smallest natural number is one. There is no greatest natural number, because no matter how large a number is, you can always add 1 to it and write the next natural number.
Let's sort it out simplest example division: divide the number 30 by the number 5 (the remainder when dividing the number 30 by the number 5 is 0), since 30 = 5. 6. So the number 30 is divisible by the number 5. The number 5 is divider the number is 30, and the number 30 is multiple number 5.
Natural number k n, if there is such a natural number m, for which the equality holds k = n . m.
Or in other words , to divide one number by another, you need to find a third number that, when multiplied by the second, gives the first
If a natural number k divisible by a natural number n, then the number k called multiples of the number,
number n — divisor of a number k.
The numbers 1, 2, 3, 6, 10, 15, 30 are also divisors of 30, and 30 is a multiple of each of these numbers. Note that the number 30 is not divisible by, for example, the number 7. Therefore, the number 7 is not a divisor of the number 30, and the number 30 is not a multiple of the number 7.
After performing division operations they say: “Number k divisible by a number n", "Number n is a divisor of a number k", "Number k multiple of the number n", "Number k is a multiple of the number n».
It is easy to write down all the divisors of the number 6. These are the numbers 1, 2, 3 and 6. Is it possible to list all the numbers that are multiples of the number 6? The numbers 6. 1, 6. 2, 6. 3, 6. 4, 6. 5, etc. are multiples of the number 6. We find that there are infinitely many numbers that are multiples of the number 6. Therefore, it is impossible to list them all.
In general, for any natural number k each of the numbers
k . 1, k . 2, k . 3, k . 4 , ...
is a multiple of the number k.
Least divisor any natural number k is the number 1, and greatest divisor- the number itself k.
Among the numbers that are multiples of k, there is no greatest, but the smallest is - this is the number itself k.
Each of the numbers 21 and 36 is divisible by the number 3, and their sum, the number 57, is also divisible by the number 3. In general, if each of the numbers k And n divisible by a number m, then the sum k+n is also divisible by a number m.
Each of the numbers 4 and 8 is not shares is an integer by the number 3, and their sum, the number 12, is not evenly divisible by the number 3. Each of the numbers 9 and 7 is not evenly divisible by the number 5, and their sum, the number 16, is not evenly divisible by the number 5. In general, if neither the number k, no number n are not evenly divisible by a number m, then the sum k + n may or may not be divisible by a whole number m.
The number 35 is divisible by the number 7 without a remainder, but the number 17 is not divisible by the number 7. The sum 35 + 17 is also not divisible by the number 7. In general, if the number k divisible by a number m and number n is not divisible by a number m, then the sum k + n is not divisible by a number m.
Regional research conference for schoolchildren of the Lakhdenpokh municipal district
"Step into the Future"
Mathematics project on the topic:
Completed by: Galkina Natalya
7th grade student
MKOU "Elisenvaara Secondary School"
Head: Vasilyeva
Larisa Vladimirovna
mathematic teacher
MKOU "Elisenvaara Secondary School"
The divisibility of numbers is determined by the last digit(s) 5 – 6 pages.
The divisibility of numbers is determined by the sum of the digits of the number: 6 pages.
The divisibility of numbers is determined after performing some actions on the digits of the number 6 - 9 pages.
To determine the divisibility of a number, other signs are used 9 – 10 pages.
Introduction 3 pages
From the history of mathematics 4 pages.
Basic concepts 4 pages.
Classification of signs of divisibility: 5 pages.
Application of divisibility criteria in practice 10 – 11 pages.
Conclusion 11 pages
Bibliography 12 pages.
Introduction
The relevance of research: Signs of divisibility have always interested scientists of different times and peoples. When studying the topic “Signs of divisibility of numbers by 2, 3, 5, 9, 10” in mathematics lessons, I became interested in studying numbers for divisibility. It was assumed that if it is possible to determine the divisibility of numbers by these numbers, then there must be signs by which one can determine the divisibility natural numbers and for other numbers. In some cases, in order to find out whether any natural number is divisible a to a natural number b without a remainder, it is not necessary to divide these numbers. It is enough to know some signs of divisibility.
Hypothesis– if there are signs of the divisibility of natural numbers by 2, 3, 5, 9 and 10, then there are other signs by which the divisibility of natural numbers can be determined.
Purpose of the study – supplement the already known signs of divisibility of natural numbers as a whole, studied at school and systematize these signs of divisibility.
To achieve this goal, it is necessary to solve the following tasks:
Independently investigate the divisibility of numbers.
Study additional literature in order to become familiar with other signs of divisibility.
Combine and summarize features from different sources.
Draw a conclusion.
Object of study– study of all possible signs of divisibility.
Subject of study– signs of divisibility.
Research methods– collection of material, data processing, comparison, analysis, synthesis.
Novelty: During the course of the project, I expanded my knowledge about the signs of divisibility of natural numbers.
From the history of mathematics
Blaise Pascal(born in 1623) - one of the most famous people in the history of mankind. Pascalumer, when he was 39 years old, but despite such short life, went down in history as an outstanding mathematician, physicist, philosopher and writer. The unit of pressure (pascal) and a very popular programming language today are named after him. Blaise Pascal found a common
Pascal's test is a method that allows you to obtain tests for divisibility by any number. A kind of “universal sign of divisibility”.
Pascal's divisibility test: Natural number A will be divided by another natural number b only if the sum of the products of the digits of the number A into the corresponding remainders obtained by dividing the digit units by the number b, is divided by this number.
For example : the number 2814 is divisible by 7, since 2 6 + 8 2 + 1 3 + 4 = 35 is divisible by 7. (Here 6 is the remainder of the division of 1000 by 7, 2 is the remainder of dividing 100 by 7 and 3 is the remainder from dividing 10 by 7).
Basic Concepts
Let's remember some mathematical concepts that we will need when studying this topic.
Divisibility test is a rule by which, without performing division, you can determine whether one number is divisible by another.
Divider natural number A name the natural number to which A divided without remainder.
Simple are called natural numbers that have no other natural distinct divisors except one and themselves.
Composite are numbers that have natural divisors other than 1 and themselves.
Signs of divisibility
All the signs of divisibility of natural numbers that I considered in this work can be divided into 4 groups:
Let's take a closer look at each of these groups.
The divisibility of numbers is determined by the last digit(s)
The first group of signs of divisibility of natural numbers that I considered includes signs of divisibility by 2, 4, 5, 8, 20, 25, 50, 125 and digit units 10, 100, etc.
Test for divisibility by 2: A number is divisible by 2 when the last digit of that number is divisible by 2 (i.e. the last digit is an even number).
For example: 32217864 : 2
Test for divisibility by 4 : A number is divisible by 4 when its last two digits are zeros, or when the two-digit number formed by its last two digits is divisible by 4.
For example, 35324 : 4; 6600 : 4
Divisibility test by 5 : A number is divisible by 5 when its last digit is 5 or 0.
For example: 36780 : 5 or 12326 5 : 5
Test for divisibility by 8: a number is divisible by 8 when it is divisible by 8 three digit number, formed from the last three digits of this number.
For example: 432240 : 8
Test for divisibility by 20: a number is divisible by 20 when the number formed by the last two digits is divisible by 20. (Another formulation: a number is divisible by 20 when the last digit of the number is 0 and the penultimate digit is even).
For example: 59640 : 20
Test for divisibility by 25: Numbers whose last two digits are zeros or form a number that is divisible by 25 are divisible by 25.
For example: 667975 : 25 or 77689 00 : 25
Test for divisibility by 50: A number is divisible by 50 when the number formed by its two lowest decimal digits is divisible by 50.
For example: 564350 :50 or 5543 00 :50
Divisibility test by 125: A number is divisible by 125 if its last three digits are zeros or form a number that is divisible by 125.
For example: 32157000 :125 or 3216 250 :125
Those natural numbers whose number of zeros is greater than or equal to the number of zeros of the digit unit are divided into a digit unit.
For example, 12,000 is divisible by 10, 100 and 1000.
The divisibility of numbers is determined by the sum of the digits of the number
This group of signs of divisibility of natural numbers includes the signs of divisibility by 3, 9, 11 that I considered.
Test for divisibility by 3: A number is divisible by 3 if its sum of digits is divisible by 3.
For example: 5421: 3 tk. 5+4+2+1=12, (12:3)
Test for divisibility by 9: A number is divisible by 9 if its sum of digits is divisible by 9.
For example: 653022: 9 tk. 6+5+3+0+2+2=18, (18:9)
Test for divisibility by 11: Those numbers are divisible by 11 if the sum of the digits in odd places is either equal to the sum of the digits in even places or differs from it by a multiple of 11.
For example: 865948732:11 because 8+5+4+7+2=26 and 6+9+8+3=26 (26=26); 815248742:11 because 8+5+4+7+2=26 and 1+2+8+4=15, 26-15=11, (11:11)
The divisibility of numbers is determined after performing some actions on the digits of this number
This group of signs of divisibility of natural numbers includes signs of divisibility by: 6, 7, 11, 13,17, 19, 23, 27, 29, 31, 33, 37, 41, 59, 79, 101
Test for divisibility by 6:
Sign 1: A number is divisible by 6 when the result of subtracting twice the number of hundreds from the number after the hundreds is divisible by 6.
For example, 138: 6 because 1·2=2, 38 – 2=36, (36:6); 744:6 because 44 – 7·2=30, (30:6)
Sign 2: A number is divisible by 6 if and only if quadruple the number of tens added to the number of units is divisible by 6.
For example, 768:6 because 76·4+8=312, 31·4+2=126, 12·4+6=54 (54:6)
Divisibility by 7:
Sign 1: number is divisible by 7 when triple the number of tens added to the number of units is divisible by 7.
For example, number 154:7, because 15 3 + 4 = 49 (49:7) is divided by 7
Sign 2: a number is divisible by 7 when the modulus of the algebraic sum of numbers forming odd groups of three digits (starting with ones), taken with the “+” sign, and even numbers with the “-” sign is divisible by 7.
For example, 138689257:7, because ǀ138-689+257ǀ=294 (294:7)
Divisibility by 11:
Sign 1: A number is divisible by 11 when the modulus of the difference between the sum of the digits occupying odd positions and the sum of the digits occupying even positions is divisible by 11.
For example, 9163627:11, because ǀ(9+6+6+7)-(1+3+2)ǀ=22 (22:11)
Sign 2: a number is divisible by 11 when the sum of numbers forming groups of two digits (starting with ones) is divisible by 11.
For example, 103785:11, because 10+37+85=132 and 01+32=33 (33:11)
Divisibility by 13:
Sign 1: A number is divisible by 13 when the sum of the tens number plus four times the ones is divisible by 13.
For example, 845:13, because 84+5·4=104, 10+4·4=26 (26:13)
Sign 2: A number is divisible by 13 when the difference between the number of tens and nine times the number of ones is divisible by 13.
For example, 845:13, because 84-5 9=39 (39:13)
Test for divisibility by 17: a number is divisible by 17 when the modulus of the difference between the number of tens and five times the number of ones is divisible by 17.
For example, 221:17, because ǀ22-5·1ǀ=17
Signs of divisibility by 19: A number is divisible by 19 when the number of tens added to twice the number of units is divisible by 19.
For example, 646:19, because 64+6·2=76, 7+2·6=19, (19:19)
Tests for divisibility by 23:
Sign 1: A number is divisible by 23 when the hundreds number added to triple the number formed by the last two digits is divisible by 23.
For example, 28842:23, because 288+3·42=414, 4+3·14=46 (46:23)
Sign 2: number is divisible by 23 when the number of tens added to seven times the number of ones is divisible by 23.
For example, 391:23, because 3 9+7 1=46 (46:23)
Sign 3: number is divisible by 23 when the number of hundreds added to seven times the number of tens and triple the number of units is divisible by 23.
For example, 391:23, because 3+7·9+3·1=69 (69:23)
Test for divisibility by 27: a number is divisible by 27 when the sum of numbers forming groups of three digits (starting with ones) is divisible by 27.
For example, 2705427:27 because 427+705+2=1134, 134+1=135, (135:27)
Test for divisibility by 29: A number is divisible by 29 when the number of tens added to three times the number of units is divisible by 29.
For example, 261:29, because 26+3·1=29 (29:29)
Test for divisibility by 31: A number is divisible by 31 when the modulus of the difference between the number of tens and three times the number of ones is divisible by 31.
For example, 217:31, because ǀ21-3·7ǀ= 0, (0:31)
Tests for divisibility by 33: If the sum made up by dividing a number from right to left into groups of two digits is divisible by 33, then the number is divisible by 33.
For example, 396:33, because 96+3=99 (99:33)
Divisibility criteria by 37:
Sign 1: a number is divisible by 37 when, when dividing the number into groups of three digits (starting with ones), the sum of these groups is a multiple of 37.
For example, number 100048:37, because 100+048=148, (148:37)
Sign 2: a number is divisible by 37 when the modulus of triple the number of hundreds added to quadruple the number of tens minus the number of units multiplied by seven is divided by 37.
For example, the number is 481:37, since it is divisible by 37ǀ3·4+4·8-7·1ǀ=37
Divisibility criteria by 41:
Sign 1: A number is divisible by 41 when the modulus of the difference between the number of tens and four times the number of units is divisible by 41.
For example, 369:41, because ǀ36-4·9ǀ=0, (0:41)
Sign 2: To check whether a number is divisible by 41, it should be divided from right to left into groups of 5 digits each. Then in each group, multiply the first digit on the right by 1, multiply the second digit by 10, third by 18, fourth by 16, fifth by 37 and add all the resulting products. If the resultwill be divisible by 41, then the number itself will be divisible by 41.
Test for divisibility by 59: A number is divisible by 59 when the number of tens added to the number of ones multiplied by 6 is divisible by 59.
For example, 767:59, because 76+7·6=118, 11+8·6=59, (59:59)
Test for divisibility by 79: A number is divisible by 79 when the number of tens added to the number of ones multiplied by 8 is divisible by 79.
For example, 711:79, because 71+8·1=79, (79:79)
Divisibility test by 99: A number is divisible by 99 when the sum of numbers that form groups of two digits (starting with ones) is divisible by 99.
For example, 12573:99, because 1+25+73=99, (99:99)
Divisibility test by 101: a number is divisible by 101 when the modulus of the algebraic sum of numbers forming odd groups of two digits (starting with ones), taken with the “+” sign, and even numbers with the “–” sign is divisible by 101.
For example
To determine the divisibility of a number, other divisibility criteria are used
This group of signs of divisibility of natural numbers includes signs of divisibility by: 6, 12, 14, 15, 27, 30, 60, etc. These are all composite numbers. Divisibility criteria for composite numbers are based on divisibility criteria for prime numbers, into which any composite number can be decomposed.
Test for divisibility by 6:
Sign 1: A number is divisible by 6 when it is divisible by both 2 and 3, that is, if it is even and the sum of its digits is divisible by 3.
For example, 768:6, because 7+6+8=21 (21:3) and the last digit in the number 768 is even.
Divisibility test by 12: A number is divisible by 12 when it is divisible by 3 and 4 at the same time.
For example, 408:12, because 4+0+8=12 (12:3) and the last two digits are divisible by 4 (08:4)
Test for divisibility by 14: A number is divisible by 14 when it is divisible by 2 and 7.
For example, the number 45612:14 because it is divisible by both 2 and 7, which means it is divisible by 14.
Test for divisibility by 15: A number is divisible by 15 when it is divisible by 3 and 5.
For example, 1146795:15 because This number is divisible by both 3 and 5.
Tests for divisibility by 27: A number is divisible by 27 when it is divisible by 3 and 9.
For example, 511704:27 because 5+1+1+7+0+4=18, (18:3 and 18:9)
Signs of divisibility by 30: A number is divisible by 30 when it ends in 0 and the sum of all digits is divisible by 3.
For example, 510:30 because 5+1+0=6 (6:3) and in the number 510 (last digit 0)
Signs of divisibility by 60: In order for a number to be divisible by 60, it is necessary and sufficient that it be divisible by 4, 3, or 5.
For example, 1620:60 because 1+6+2+0=9 (9:3), the number 1620 ends with 0, i.e. is divisible by 5 and 1620: 4 because last two digits 20:4
The work has practical application. It can be used by schoolchildren and adults when solving real situations; teachers, both when conducting mathematics lessons and in elective courses and additional classes for repetition.
This study will be useful for students when self-training for final and entrance exams. It will also be useful for students whose goal is high places at city Olympiads.
Task No. 1 . Is it possible, using only the numbers 3 and 4, to write:
a number that is divisible by 10;
even number;
a number that is a multiple of 5;
odd number
Problem No. 2
Write some nine-digit number that has no repeating digits (all digits are different) and is divisible by 1 without a remainder.
Write the largest of these numbers.
Write the smallest of these numbers.
Answer: 987652413; 102347586
Problem No. 3
Find the largest four-digit number, all of whose digits are different and which is divisible by 2, 5, 9, 11.
Answer: 8910
Problem No. 4
Olya came up with a simple three-digit number, all of whose digits are different. What digit can it end in if its last digit is equal to the sum of the first two. Give examples of such numbers.
Answer: only by 7. There are 4 numbers that satisfy the conditions of the problem: 167, 257, 347, 527
Problem No. 5
There are 70 students in the two classes together. In one class, 7/17 students did not show up for classes, and in another, 2/9 received excellent grades in mathematics. How many students are in each class?
Solution: In the first of these classes there could be: 17, 34, 51... - numbers that are multiples of 17. In the second class: 9, 18, 27, 36, 45, 54... - numbers that are multiples of 9. We need to choose 1 number from the first sequence , and 2 is a number from the second so that they add up to 70. Moreover, in these sequences only a small number of terms can express the possible number of children in the class. This consideration significantly limits the selection of options. The only possible option was the pair (34, 36).
Problem No. 6
In 9th grade for test 1/7 students received A's, 1/3 - B's, ½ - C's. The rest of the work turned out to be unsatisfactory. How many such jobs were there?
Solution: The solution to the problem must be a number that is a multiple of the numbers: 7, 3, 2. Let's first find the smallest of these numbers. LCM (7, 3, 2) = 42. You can create an expression according to the conditions of the problem: 42 – (42: 7 + 42: 3 + 42: 2) = 1 – 1 unsuccessful. Mathematical relationship problems assume that the number of students in the class is 84, 126, etc. Human. But common sense suggests that the most acceptable answer is the number 42.
Answer: 1 job.
Conclusion:
As a result of this work, I learned that in addition to the signs of divisibility by 2, 3, 5, 9 and 10 that I know, there are also other signs of divisibility of natural numbers. The knowledge gained significantly speeds up the solution of many problems. And I can use this knowledge in my educational activities, both in mathematics lessons and in extracurricular activities. It should also be noted that the formulations of some divisibility criteria are complex. Maybe that's why they are not studied in school. I expect to continue to work on studying the signs of divisibility of natural numbers in the future.
encyclopedic Dictionary young mathematician. Savin A.P. Moscow "Pedagogy" 1989.
Mathematics. Additional materials for mathematics lessons, grades 5-11. Ryazanovsky A.R., Zaitsev E.A. Moscow “Bustard” 2002.
Behind the pages of a mathematics textbook. Vilenkin N.Ya., Depman I.Ya. M.: Education, 1989.
Extracurricular work in mathematics in grades 6-8. Moscow. “Enlightenment” 1984 V. A. Gusev, A. I. Orlov, A. L. Rosenthal.
“1001 questions and answers. Big book of knowledge" Moscow. "World of Books" 2004.
Optional course in mathematics. Nikolskaya I.L. - Moscow. Enlightenment 1991.
Olympiad problems in mathematics and methods for solving them. Farkov A.V. - Moscow. 2003
Internet resources.
View presentation content
“Signs of divisibility of natural numbers”
Regional research conference for schoolchildren
Lakhdenpokh municipal district “Step into the future”
“Signs of divisibility of natural numbers”
Completed by: Galkina Natalya
7th grade student
MKOU "Elisenvaara Secondary School"
Head: Vasilyeva Larisa Vladimirovna
mathematics teacher at MKOU "Elisenvaarskaya" Secondary School"
2014
The relevance of research : Signs of divisibility have always interested scientists of different times and peoples. When studying the topic “Signs of divisibility of numbers by 2, 3, 5, 9, 10” in mathematics lessons, I became interested in studying numbers for divisibility. It was assumed that if it is possible to determine the divisibility of numbers by these numbers, then there must be signs by which one can determine the divisibility of natural numbers by other numbers. In some cases, in order to find out whether any natural number is divisible a to a natural number b without a remainder, it is not necessary to divide these numbers. It is enough to know some signs of divisibility. Hypothesis – if there are signs of the divisibility of natural numbers by 2, 3, 5, 9 and 10, then there are other signs by which the divisibility of natural numbers can be determined. Purpose of the study – supplement the already known signs of divisibility of natural numbers as a whole, studied at school and systematize these signs of divisibility. To achieve this goal, it is necessary to solve the following tasks:
- Independently investigate the divisibility of numbers.
- Study additional literature in order to become familiar with other signs of divisibility.
- Combine and summarize features from different sources.
- Draw a conclusion. Object of study – divisibility of natural numbers. Subject of study – signs of divisibility. Research methods – collection of material, data processing, comparison, analysis, generalization. Novelty : During the project I expanded my knowledge on criteria for the divisibility of natural numbers.
From the history of mathematics
Blaise Pascal (born 1623) - one of the most famous people in human history. Pascal died when he was 39 years old, but despite such a short life, he went down in history as an outstanding mathematician, physicist, philosopher and writer. The unit of pressure (pascal) and a very popular programming language today are named after him. Blaise Pascal found a common an algorithm for finding signs of divisibility of any integer by any other integer.
Pascal's test is a method that allows you to obtain tests for divisibility by any number. A kind of “universal sign of divisibility”.
Pascal's divisibility test: A natural number a will be divided by another natural number b only if the sum of the products of the digits of the number a by the corresponding remainders obtained by dividing the digit units by the number b is divisible by this number.
For example : the number 2814 is divisible by 7, since 2 6 + 8 2 + 1 3 + 4 = 35 is divisible by 7. (Here 6 is the remainder of the division of 1000 by 7, 2 is the remainder of dividing 100 by 7 and 3 is the remainder from dividing 10 by 7).
Basic Concepts
Let's remember some mathematical concepts that we will need when studying this topic:
- Divisibility test is a rule by which, without performing division, you can determine whether one number is divisible by another.
- Divider natural number A call a natural number b , to which A divided without remainder.
- Simple are called natural numbers that have no other natural distinct divisors except one and themselves.
- Composite are numbers that have natural divisors other than 1 and themselves.
Signs of divisibility
All the signs of divisibility of natural numbers that I considered in this work can be divided into 4 groups:
I
- I . The divisibility of numbers is determined by the last digit(s)
The first group of signs of divisibility of natural numbers that I considered includes signs of divisibility by 2, 4, 5, 8, 20, 25, 50, 125 and digit units 10, 100, etc.
- Test for divisibility by 2 : A number is divisible by 2 when the last digit of that number is divisible by 2 (i.e. the last digit is an even number).
For example : 3221786 4 : 2
- Test for divisibility by 4 : A number is divisible by 4 when its last two digits are zeros, or when the two-digit number formed by its last two digits is divisible by 4.
For example: 353 24 : 4; 66 00 : 4
- Divisibility test by 5 : A number is divisible by 5 when its last digit is 5 or 0.
For example: 3678 0 : 5 or 12326 5 : 5
- Test for divisibility by 8: A number is divisible by 8 when a three-digit number formed from the last three digits of that number is divisible by 8.
For example: 432 240 : 8
- Test for divisibility by 20: a number is divisible by 20 when the number formed by two last numbers, divisible by 20. (Another formulation: the number is divisible by 20 when the last digit of the number is 0, and the second to last digit is even).
For example: 596 40 : 20
- Test for divisibility by 25: Numbers whose last two digits are zeros or form a number that is divisible by 25 are divisible by 25.
For example: 6679 75 : 25 or 77689 00 : 25
- Test for divisibility by 50: A number is divisible by 50 when the number formed by its two lowest decimal digits is divisible by 50.
For example : 5643 50 : 50 or 5543 00 : 50
- Divisibility test by 125: A number is divisible by 125 if its last three digits are zeros or form a number that is divisible by 125.
For example: 32157 000 : 125 or 3216 250 : 125
- Signs of divisibility by digit unit 10, 100, 1000, etc.: Those natural numbers whose number of zeros is greater than or equal to the number of zeros of the digit unit are divided into a digit unit.
For example, 12,000 is divisible by 10, 100 and 1000
II
- II . The divisibility of numbers is determined by the sum of the digits of the number
This group of signs of divisibility of natural numbers includes the signs of divisibility by 3, 9, 11 that I considered.
- Test for divisibility by 3: A number is divisible by 3 if its sum of digits is divisible by 3.
For example: 5421: 3 tk. 5+4+2+1=12, (12:3)
- Test for divisibility by 9: A number is divisible by 9 if its sum of digits is divisible by 9.
For example: 653022: 9 because 6+5+3+0+2+2=18, (18:9)
- Test for divisibility by 11: Those numbers are divisible by 11 if the sum of the digits in odd places is either equal to the sum of the digits in even places or differs from it by a multiple of 11.
For example: 865948732:11 because 8+5+4+7+2=26 and 6+9+8+3=26 (26=26); 815248742:11 because 8+5+4+7+2=26 and 1+2+8+4=15, 26-15=11, (11:11)
III . The divisibility of numbers is determined after performing some actions
above the digits of this number
This group of signs of divisibility of natural numbers includes signs of divisibility by: 6, 7, 11, 13,17, 19, 23, 27, 29, 31, 33, 37, 41, 59, 79, 99, 101
Test for divisibility by 6:
- Sign 1: a number is divisible by 6 when the result of subtracting twice the number of hundreds from the number after the hundreds is divisible by 6.
For example: 138: 6 because 1·2=2, 38 – 2=36, (36:6); 744:6 because 44 – 7·2=30, (30:6)
- Sign 2: a number is divisible by 6 if and only if the quadruple number of tens added to the number of ones is divisible by 6.
For example: 768:6 because 76·4+8=312, 31·4+2=126, 12·4+6=54 (54:6)
Divisibility by 7:
- Sign 1: a number is divisible by 7 when triple the number of tens added to the number of ones is divisible by 7.
For example: the number 154:7, because 15 3 + 4 = 49 (49:7) is divided by 7
- Sign 2: a number is divisible by 7 when the modulus of the algebraic sum of numbers forming odd groups of three digits (starting with ones), taken with the “+” sign, and even numbers with the “-” sign is divisible by 7.
For example, 138689257:7, because ǀ138-689+257ǀ=294 (294:7)
Divisibility by 11:
- Sign 1: a number is divisible by 11 when the modulus of the difference between the sum of the digits occupying odd positions and the sum of the digits occupying even positions is divisible by 11.
For example, 9163627:11, because ǀ(9+6+6+7)-(1+3+2)ǀ=22 (22:11)
- Sign 2: a number is divisible by 11 when the sum of numbers forming groups of two digits (starting with ones) is divisible by 11.
For example, 103785:11, because 10+37+85=132 and 01+32=33 (33:11)
Divisibility by 13:
- Sign 1: a number is divisible by 13 when the sum of the number of tens and quadruple the number of ones is divisible by 13
For example, 845:13, because 84+5·4=104, 10+4·4=26 (26:13)
- Sign 2: a number is divisible by 13 when the difference between the number of tens and nine times the number of ones is divisible by 13.
For example, 845:13, because 84-5 9=39 (39:13)
Test for divisibility by 17: a number is divisible by 17 when the modulus of the difference between the number of tens and five times the number of ones is divisible by 17.
For example, 221:17, because ǀ22-5·1ǀ=17
Signs of divisibility by 19: a number is divisible by 19 when the number is tens, with false with double the number of units, divisible by 19.
For example, 646:19, because 64+6·2=76, 7+2·6=19, (19:19)
Tests for divisibility by 23:
- Sign 1: a number is divisible by 23 when the number of hundreds added to triple the number formed by the last two digits is divisible by 23.
For example, 28842:23, because 288+3·42=414, 4+3·14=46 (46:23)
- Sign 2: a number is divisible by 23 when the number of tens added to seven times the number of units is divisible by 23.
For example, 391:23, because 39+7·1=46 (46:23)
- Sign 3: a number is divisible by 23 when the number of hundreds, added to seven times the number of tens and triple the number of units, divisible by 23.
For example, 391:23, because 3+7·9+3·1=69 (69:23)
Test for divisibility by 27: a number is divisible by 27 when the sum of numbers forming groups of three digits (starting with ones) is divisible by 27.
For example, 2705427:27 because 427+705+2=1134, 134+1=135, (135:27)
Test for divisibility by 29: a number is divisible by 29 when the number of tens added to three times the number of ones is divisible by 29
For example, 261:29, because 26+3·1=29 (29:29)
Test for divisibility by 31: a number is divisible by 31 when the modulus of the difference of the number of tens and three times the number of units is divided by 31.
For example, 217:31, because ǀ21-3·7ǀ= 0, (0:31)
Tests for divisibility by 33: If the sum made up by dividing a number from right to left into groups of two digits is divisible by 33, then the number is divisible by 33.
For example, 396:33, because 96+3=99 (99:33)
Divisibility criteria by 37:
- Sign 1 : a number is divisible by 37 when, when dividing the number into groups of three digits (starting with ones), the sum of these groups is a multiple of 37.
For example , number 100048:37, because 100+048=148, (148:37)
- Sign 2: a number is divisible by 37 when the module of triple the number of hundreds, added to quadruple the number of tens, minus the number of units multiplied by seven, is divisible by 37.
For example, the number 481:37, since ǀ3·4+4·8-7·1ǀ=37 is divisible by 37
Divisibility criteria by 41:
- Sign 1: a number is divisible by 41 when the modulus of the difference between the number of tens and four times the number of ones is divisible by 41.
For example, 369:41, because ǀ36-4·9ǀ=0, (0:41)
- Sign 2: to check whether a number is divisible by 41, it should be divided from right to left into groups of 5 digits each. Then in each group, multiply the first digit on the right by 1, multiply the second digit by 10, third by 18, fourth by 16, fifth by 37 and add all the resulting products. If the result is divisible by 41, then the number itself will be divisible by 41.
Test for divisibility by 59: A number is divisible by 59 when the number of tens added to the number of ones multiplied by 6 is divisible by 59.
For example, 767:59, because 76+7·6=118, 11+8·6=59, (59:59)
Test for divisibility by 79: A number is divisible by 79 when the number of tens added to the number of ones multiplied by 8 is divisible by 79.
For example, 711:79, because 71+8·1=79, (79:79)
Divisibility test by 99: A number is divisible by 99 when the sum of numbers that form groups of two digits (starting with ones) is divisible by 99.
For example, 12573:99, because 1+25+73=99, (99:99)
Divisibility test by 101: a number is divisible by 101 when the modulus of the algebraic sum of numbers forming odd groups of two digits (starting with ones), taken with the “+” sign, and even numbers with the “–” sign is divisible by 101.
For example, 590547:101, because ǀ59-5+47ǀ=101, (101:101)
IV . To determine the divisibility of a number, other divisibility criteria are used
This group of signs of divisibility of natural numbers includes signs of divisibility by: 6, 12, 14, 15, 27, 30, 60, etc. These are all composite numbers. Divisibility criteria for composite numbers are based on divisibility criteria for prime numbers, into which any composite number can be decomposed.
Test for divisibility by 6: A number is divisible by 6 when it is divisible by both 2 and 3, that is, if it is even and the sum of its digits is divisible by 3.
For example, 768:6, because 7+6+8=21 (21:3) and the last digit in the number 768 is even.
Divisibility test by 12 : A number is divisible by 12 when it is divisible by 3 and 4 at the same time.
For example, 408:12, because 4+0+8=12 (12:3) and the last two digits are divisible by 4 (08:4)
Test for divisibility by 14: A number is divisible by 14 when it is divisible by 2 and 7.
For example, the number 45612:14 because it is divisible by both 2 and 7, which means it is divisible by 14
Test for divisibility by 15: A number is divisible by 15 when it is divisible by 3 and 5.
For example, 1146795:15 because this number is divisible by both 3 and 5
Tests for divisibility by 27: A number is divisible by 27 when it is divisible by 3 and 9. For example, 511704:27 because 5+1+1+7+0+4=18, (18:3 and 18:9)
Signs of divisibility by 30: A number is divisible by 30 when it ends in 0 and the sum of all digits is divisible by 3.
For example, 510:30 because 5+1+0=6 (6:3) and in the number 510 (last digit 0)
Signs of divisibility by 60: In order for a number to be divisible by 60, it is necessary and sufficient that it be divisible by 4, 3, or 5.
For example, 1620:60 because 1+6+2+0=9 (9:3), the number 1620 ends with 0, i.e. is divisible by 5 and 1620: 4 because last two digits 20:4
Application of divisibility criteria in practice
The work has practical application. It can be used by schoolchildren and adults when solving real situations; teachers, both during mathematics lessons and in elective courses and additional revision classes.
This study will be useful for students in their independent preparation for final and entrance exams. It will also be useful for students whose goal is high places at city Olympiads.
Task No. 1 . Is it possible, using only the numbers 3 and 4, to write:
- a number that is divisible by 10;
- even number;
- a number that is a multiple of 5;
- odd number
Problem No. 3 : Find the largest four-digit number, all of whose digits are different and which is divisible by 2, 5, 9, 11.
Answer: 8910
Task #4: Olya came up with a simple three-digit number, all of whose digits are different. What digit can it end in if its last digit is equal to the sum of the first two. Give examples of such numbers.
Answer: only by 7. There are 4 numbers that satisfy the conditions of the problem: 167, 257, 347, 527
Problem No. 5 : There are 70 students in two classes together. In one class, 7/17 students did not show up for classes, and in another, 2/9 received excellent grades in mathematics. How many students are in each class?
Solution: In the first of these classes there could be: 17, 34, 51... - numbers that are multiples of 17. In the second class: 9, 18, 27, 36, 45, 54... - numbers that are multiples of 9. We need to choose 1 number from the first sequence , and 2 is a number from the second so that they add up to 70. Moreover, in these sequences only a small number of terms can express the possible number of children in the class. This consideration significantly limits the selection of options. The only possible option was the pair (34, 36).
Problem No. 6 : In the 9th grade, 1/7 students received A’s for the test, 1/3 received fours, ½ - threes. The rest of the work turned out to be unsatisfactory. How many such works were there?
Solution: The solution to the problem must be a number that is a multiple of the numbers: 7, 3, 2. Let’s find first the smallest of these numbers. LCM (7, 3, 2) = 42. You can make an expression according to the conditions of the problem: 42 – (42: 7 + 42: 3 + 42: 2) = 1 – 1 unsuccessful. Mathematical relationship problems assume that the number students in class 84, 126, etc. Human. But for reasons of common sense It follows that the most acceptable answer is the number 42.
Answer: 1 job.
Conclusion:
As a result of this work, I learned that in addition to the signs of divisibility by 2, 3, 5, 9 and 10 that I know, there are also other signs of divisibility of natural numbers. The knowledge gained significantly speeds up the solution of many problems. And I will be able to use this knowledge in my educational activities, both in mathematics lessons and in extracurricular activities. It should also be noted that the formulations of some divisibility criteria are complex. Maybe that's why they are not studied in school. I expect to continue to work on studying the signs of divisibility of natural numbers in the future.
- Encyclopedic dictionary of a young mathematician. Savin A.P. Moscow "Pedagogy" 1989.
- Mathematics. Additional materials for mathematics lessons, grades 5-11. Ryazanovsky A.R., Zaitsev E.A. Moscow “Bustard” 2002.
- Behind the pages of a mathematics textbook. Vilenkin N.Ya., Depman I.Ya. M.: Education, 1989.
- Extracurricular work in mathematics in grades 6-8. Moscow. “Enlightenment” 1984 V. A. Gusev, A. I. Orlov, A. L. Rosenthal.
- “1001 questions and answers. Big book of knowledge" Moscow. "World of Books" 2004.
- Optional course in mathematics. Nikolskaya I.L. - Moscow. Enlightenment 1991.
- Olympiad problems in mathematics and methods for solving them. Farkov A.V. - Moscow. 2003
- Internet resources.
Integers
A set of natural numbers used for counting or transferring.
Formally, the set of natural numbers can be defined using the Peano axiom system.
WITHPeano axiom system
1. Unit 
- a natural number that does not follow any number.
2. For any natural number 
exists singular 
which immediately follows .
3. Every natural number 
immediately follows only one number.
4. If some set 
contains and together with each natural number contains the number immediately following it then 
(axiom of induction).
Operations on a set
Multiplication
|
Subtraction :
Subtraction Properties: If 
That
If 
That
Divisibility of natural numbers
Division
:
divided by 
such that 
Propertiesoperations:
1. If 
are divided into 
That 
divided by 
2. If 
And 
are divided into 
That 
divided by 
3. If 
And 
are divisible by that is divisible by
4. If divisible by then 
divided by
5. If 
are divisible by a 
are not divided into this and that 
not divisible by
6. If or 
divided by that 
divided by
7. If divisible by 
then it is divided by 
and is divided by
Theoremabout division with remainder For any natural numbers 
there are only positive numbers 
such that 
and 
Proof. Let 
Consider the following algorithm:
| If If We continue the subtraction process until the remainder is less than the number There is a number |
| Let's add up all the lines of this algorithm and get the required expression, where
|
then let's do another subtraction
such that 
We will prove the uniqueness of the representation by contradiction.
Suppose there are two representations
And 
Subtract one expression from the other and 
The last equality in integers is possible only in the case since 
at 
□
Corollary 1. Any natural number can be represented as: 
or or
Corollary 2. If 
consecutive natural numbers, then one of them is divisible by 
Corollary 3. If 
two consecutive even numbers, then one of them is divisible by 
Definition.
Natural number 
is called prime if it has no divisors other than one and itself.
Consequence4.
Every prime number has the form 
or 
Indeed, any number can be represented in the form; however, all numbers in this series, except 
are definitely composite. □
Consequence5
. If 
prime number then 
divided by 
Really, 
three consecutive natural numbers, and 
even, and 
odd prime. Therefore, one of the even numbers 
And 
is divisible by 4, and one is also divisible by 
□
Example 2 . The following statements are true:
1. The square of an odd number when divided by 8 gives a remainder 
2. For no natural number n is the number n 2 +1 divisible by 3.
3. Using only the numbers 2, 3, 7, 8 (possibly several times), it is impossible to square a natural number.
Proof1.
Any odd number can be represented as 
or 
Let's square each of these numbers and get the required statement.
Proof 2. Every natural number can be represented as 
Then the expression 
will be equal to one of the expressions 
which are not divided into 
Proof3. Indeed, the last digit of the square of a natural number cannot end in any of these digits.
Signs of divisibility
Definition. The decimal representation of a natural number is the representation of a number in the form
Shorthand notation 
Signs of divisibility into
Approved 6 Let 
decimal representation of the number number Then:
1. The number is divisible by 
when the number 
- even;
2. The number is divisible by 
when the number is two digits 
divided by 
3. The number is divisible by 
When 
or 
4. The number is divisible by 
When 
5. The number is divisible by 
when the number is two digits 
- divided by 
6. The number is divisible by 
7. The number is divisible by 
when the sum of the digits of a number is divided by 
8. The number is divisible by 
when the sum of the digits of a number with alternating signs is divided by 
Proof. The proof of signs 1)-5) is easily obtained from decimal notation numbers Let us prove 6) and 7). Really,
It follows that if divisible (or 
then the sum of the digits of the number is also divisible by 
Let us prove 11). Let it be divisible by Let us represent the number in the form
Since all added sums are divisible by 
then the amount is also divided by □
Example 3
.
Find all five-digit numbers of the form 
, which are divisible by 45.
Proof.
Therefore, the number is divisible by 5, and its last digit is 0 or 5, i.e. 
or 
The original number is also divisible by 9, so it is divisible by 9, i.e. 
or divisible by 9, i.e. 
Answer:
Divisibility test on 
And 
Approved 7 Let the decimal representation of the number Number Number be divisible by 
when the difference between a number without the last three digits and a number made up of the last three digits is divided by 
Proof. Let's represent it in the form Since the number 
divided by and 
That 
divisible by and □
Example
4
.
Let 
Then 
is divisible by and therefore the number 
divided by 
Let 
Then 
divisible by Then the number 
divided by
Prime numbers
Sieve of Eratosthenes
(Simple algorithm for getting all prime numbers)
Algorithm. We write down all the numbers from 1 to 100 and cross out all the even ones first. Then, from the remaining ones, we cross out those divisible by 3, 5, 7, etc. As a result, only prime numbers will remain.
Euclid's theorem. The number of prime numbers is infinite.
Proof"by contradiction." Let the number of prime numbers be finite - 
Consider the number 
Question: number 
- simple or compound?
If is a composite number, then it is divisible by some prime number 
and therefore one is divided by this prime number. Contradiction.
If is a prime number, then it is greater than any prime number 
and we wrote out and numbered all the prime numbers. Again a contradiction. □
Approved 8 If a number is composite, then it has a prime divisor such that 
Proof. If is the smallest prime divisor of a composite number 
That 
Consequence. To determine whether a number is prime, you need to determine whether it has prime divisors. 
Example
5
. Let 
To check if a number is 
simple, you need to check whether it is divisible by prime numbers Answer: number 
simple.
Prime number generators
Hypothesis: All numbers of the form 
simple.
At 
- these are prime numbers 
For 
It has been proven manually and with the help of a computer that all numbers are composite.
For example, (Euler)
Hypothesis: All numbers of the form 
simple.
At 
that's true, eh 
divisible by 17.
Hypothesis: All numbers of the form 
simple.
At 
that's true, eh 
Hypothesis: All numbers of the form are prime. At 
that's true, eh 
Theorem.(Fermat method of factoring) Odd integer is not prime 
there are natural numbers such that 
Proof.
□
Example
6
.
Factor numbers into prime factors 
Example
7
.
Factor a number 
This number is divisible by 3 
Further, according to the method of selecting factors, 
Example
8
. At what integers 
simple?
Note that since 
simple, then either 
or 
Answer: 
Approved 10 Does a natural number have an odd number of divisors when it is a perfect square?
Proof. If 
divisor 
then has two different pairs of divisors 
And 
and when 
both pairs will be equal.
Example 9 . The numbers have exactly 99 divisors. Can a number have exactly 100 divisors?
Answer: no. Valid by the previous property and - perfect squares, but their work is not.
Example
10
.
Numbers 
simple. Find 
Solution. Any number can be represented as 
If 
then you get three prime numbers 
satisfying the conditions of the problem. If 
That 
composite. If 
that number 
divided by 
and if 
that number 
is divisible by Thus, in all the considered options three prime numbers cannot be obtained. Answer: 
Definition.
Number 
is called the greatest common divisor of numbers and if it divides and and is the largest of such numbers.
Designation: 
Definition
.
Numbers and are said to be relatively prime if 
Example 1
2
.
Solve the equation in natural numbers 
Solution. Let 
Therefore, the equation looks like Answer: There are no solutions.
ABOUTfundamental theorem of arithmetic
Theorem. Any natural number greater than is either a prime number or can be written as a product of prime numbers, and this product is unique up to the order of the factors.
Corollary 1. Let 
Then 
is equal to the product of all common prime factors with the smallest powers.
Corollary 2. Let 
Then 
is equal to the product of all different prime factors with the greatest powers. divided by
10. Find last digit numbers 7 2011 + 9 2011.
11. Find all natural numbers that increase by 9 times if a zero is inserted between the units digit and the tens digit.
12. To some two-digit number, one was added to the left and right. The result was a number 23 times larger than the original. Find this number.
Questions about theory or exercises can be asked to Valery Petrovich Chuvakov
chv @ uriit . ru
additional literature
1. Vilenkin N.Ya. and others. Behind the pages of a mathematics textbook. Arithmetic. Algebra. –M.: Education, 2008.
2. Sevryukov P.F. Preparation for solving Olympiad problems in mathematics. –M.: Ilexa, 2009.
3. Kanel-Belov A.Ya., Kovaldzhi A.K. How they decide non-standard tasks. –M. MCNMO, 2009.
4. Agakhanov N.A., Podlipsky O.K. Mathematical Olympiads of the Moscow region. –M.: Fizmatkniga, 2006
5. Gorbachev N.V. Collection of Olympiad problems, –M.:MCNMO, 2004
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