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.

Page navigation.

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 (let’s denote it 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 number a , as well as the divisibility of a by minus one.

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 is 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 integer −b, the opposite of 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. You can go to to the following property 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 and 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 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.

References.

• 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.

Natural numbers

Many natural numbers, used for invoicing or transfer.

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 one positive numbers
such that
and

Proof. Let
Consider the following algorithm:

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. Really, last digit 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 two-digit number
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 everything five digit numbers kind
, 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. Number prime numbers endlessly.

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 factors

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 what 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
equal to the product all common prime factors with the smallest powers.

Corollary 2. Let
Then
is equal to the product of all different prime factors with to the greatest extent. divided by

10. Find the last digit of the number 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

Further reading

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 Moscow region. –M.: Fizmatkniga, 2006

5. Gorbachev N.V. Collection of Olympiad problems, –M.:MCNMO, 2004

Regional research conference of schoolchildren of Lakhdenpokhsky 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

math teacher

MKOU "Elisenvaara Secondary School"

Introduction 3 pages

From the history of mathematics 4 pages.

Basic concepts 4 pages.

Classification of signs of divisibility: 5 pages.

1. 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.

Application of divisibility criteria in practice 10 – 11 pages.

Conclusion 11 pages

Bibliography 12 pages.

Introduction

Relevance of the study: 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– study of all possible signs of divisibility.

Subject of research– 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 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, which 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 others natural divisors except 1 and himself.

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 tens number added to number of units 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 ones 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 algebraic sum numbers forming odd groups of three digits (starting with ones), taken with a “+” sign, and even numbers with a “-” sign, 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)

Tests for divisibility 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 elective courses And additional classes for repetition.

This study will be useful for students when self-training for graduation and entrance exams. It will also be useful for students whose goal is high places at city Olympics.

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 possible quantity 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 work 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 for reasons 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 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 a math lesson for 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 activities in mathematics in grades 6-8. Moscow. “Enlightenment” 1984 V. A. Gusev, A. I. Orlov, A. L. Rosenthal.

“1001 questions and answers. Big book 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

Relevance of the study : 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 research – 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 units 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 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 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)

Tests for divisibility 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.

Educational field: natural science.

Section: "Mathematics"

Research work on the topic:

"Signs of divisibility of natural numbers"

Head: Lapko I.V.

math teacher

Introduction:

1. Facts from the history of mathematics.

2. Signs of divisibility by 2, 3, 4, 5,6,8, 9, 10.

3. Signs of divisibility of natural numbers by 7, 11, 12, 13, 14, 19, 25.50.

4. Solving problems using divisibility criteria.

6. List of used literature (sources).

Relevance: All of us at school learned the signs of divisibility, which to this day help us, without wasting unnecessary time, quickly and accurately divide this or that number. Not long ago, remembering this topic, I began to wonder if there were other signs of divisibility by natural numbers. And it was this thought that pushed me to write a research paper.
Hypothesis: If you can determine the divisibility of natural numbers by 2, 3, 5, 9, 10, then most likely there are signs by which you can determine the divisibility of natural numbers by other numbers.
Object of study: divisibility of natural numbers.

Subject of research: signs of divisibility of natural numbers.

Target: supplement the already known tests for the divisibility of natural numbers as a whole, studied at school.

Tasks:
1. Define and repeat the already studied signs of divisibility by 2, 3, 5, 9, 10.
2. ​ Study additional literature confirming the correctness of the question raised about the existence of other signs of the divisibility of natural numbers.
3. Independently check and obtain signs of divisibility of natural numbers by 4, 6, 8, 15, 25.
4. Find from additional literature signs of divisibility of natural numbers by 7, 11,12,13,14.
5.Draw a conclusion.
Novelty: During the course of the project, I expanded my knowledge about the signs of divisibility of natural numbers.

Research methods: collection of material, data processing, observation, comparison, analysis, synthesis.

1. Facts from the history of mathematics

1. Sign of divisibility- an algorithm that allows you to relatively quickly determine whether a number is a multiple of a predetermined one
The test of divisibility is a rule by which, without performing division, one can determine whether one natural number is divisible by another. Signs of divisibility have always interested scientists different countries and times. Signs of divisibility by 2, 3, 5, 9, 10 have been known since ancient times. The sign of divisibility by 2 was known to the ancient Egyptians 2 thousand years BC, and the signs of divisibility by 2, 3, 5 were described in detail by the Italian mathematician Leonardo Pisanus (Latin Leonardus Pisanus, Italian Leonardo Pisano, around 1170, Pisa - around 1250 year, ibid.) - the first major mathematician medieval Europe. He is best known by his nickname Fibonacci. The Alexandrian scientist Eratosthenes, who lived in the 3rd century BC, once thought about the same question. His method of compiling a list of prime numbers was called the "sieve of Eratosthenes". Let's say we need to find all the prime numbers up to 100. Let's write all the numbers up to 100 in a row.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100.

Leaving the number 2, cross out all the rest even numbers. The first surviving number after 2 will be 3. Now, leaving the number 3, we cross out the numbers divisible by 3. Then we cross out the numbers divisible by 5. As a result, all composite numbers will be crossed out and only prime numbers will remain: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. Using this method, you can make lists of prime numbers greater than 100.

The issues of divisibility of numbers were considered by the Pythagoreans. In number theory they carried out great job according to the typology of natural numbers. The Pythagoreans divided them into classes. Classes were distinguished: perfect numbers (number equal to the sum their own divisors, for example: 6=1+2+3), friendly numbers (each of which is equal to the sum of the other’s divisors, for example 220 and 284: 284=1+2+4+5+10+20+11+22+44 +55+110; 220=1+2+4+71+142), figured numbers (triangular number, square number), prime numbers, etc. Blaise Pascal (1623-1662) made a great contribution to the study of signs of divisibility of numbers. Young Blaise showed very early math skills, learning to count before reading. In general, his example is a classic case of childhood mathematical genius. His first mathematical treatise “Experience in Theory conic sections"he wrote at 24 years old. Around the same time, he designed a mechanical adding machine, the prototype of the adding machine. IN early period In his creative work (1640-1650), the versatile scientist found an algorithm for finding signs of divisibility of any integer by any other integer, from which all particular signs follow. Its sign is as follows: 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.
When studying this topic, you need to know the concepts of divisor, multiple, prime and composite numbers. The divisor of a natural number a is a natural number b, by which a is divided without a remainder. Often, a statement about the divisibility of a number by a number b is expressed in other equivalent words: a is a multiple of b, b is a divisor of a, b divides a. Prime numbers are natural numbers that have two divisors: 1 and the number itself. For example, the numbers 5,7,19 are prime numbers because are divisible by 1 and itself. Numbers that have more than two divisors are called composite numbers. For example, the number 14 has 4 divisors: 1, 2, 7, 14, which means it is composite.

2. Signs of divisibility

To simplify the division of natural numbers, rules for division into numbers of the first ten and numbers 11, 25 were derived, which were combined into a section on signs of divisibility of natural numbers. Below are the rules according to which the analysis of a number without dividing it by another natural number will answer the question, is a natural number a multiple of the numbers 2, 3, 4, 5, 6, 9, 10, 11, 25 and the digit unit?

Natural numbers that have digits (ending in) 2,4,6,8,0 in the first digit are called even.

Divisibility test for numbers by 2

All even natural numbers are divisible by 2, for example: 172, 94.67, 838, 1670.

For example, the number 52,738 is divisible by 2 because the last digit, 8, is even; 7691 is not divisible by 2, since 1 is an odd number; 1250 is divisible by 2 because the last digit is zero.

Divisibility test for numbers by 3

All natural numbers whose sum of digits is divisible by 3 are divisible by 3. For example:
39 (3 + 9 = 12; 12: 3 = 4);

16 734 (1 + 6 + 7 + 3 + 4 = 21; 21:3 = 7).

Examples.

The number 52632 is divisible by 9 because the sum of its digits (18) is divisible by 9.

Divisibility test for numbers by 4

All natural numbers whose last two digits are zeros or a multiple of 4 are divisible by 4.
124 (24: 4 = 6);
103 456 (56: 4 = 14).

Examples.
31,700 is divisible by 4 because it ends with two zeros;
215,634 is not divisible by 4, since the last two digits give the number 34, which is not divisible by 4;
16608 is divisible by 4 because the last two digits of 08 give the number 8, which is divisible by 4.

Divisibility test for numbers by 5

Divisibility test for numbers by 6

Those natural numbers that are divisible by 2 and 3 at the same time are divisible by 6 (all even numbers that are divisible by 3). For example: 126 (b - even, 1 + 2 + 6 = 9, 9: 3 = 3).

Divisibility test for numbers by 8

Those and only those numbers that end with three zeros or whose last three digits express a number divisible by 8 are divisible by 8. Example

The number 853,000 ends with three zeros, which means it is divisible by 8

The number 381,864 is divisible by 8 because the number formed by the last three digits of 864 is divisible by 8.

PrDivisibility sign for numbers by 9

Those natural numbers whose sum of digits is a multiple of 9 are divisible by 9. For example:
1179 (1 + 1 + 7 + 9 = 18, 18: 9 = 2).

Examples.
The number 17835 is divisible by 3 and not divisible by 9, since the sum of its digits 1 + 7 + 8 + 3 + 5 = 24 is divisible by 3 and not divisible by 9.
The number 105,499 is not divisible by either 3 or 9, since the sum of its digits (29) is not divisible by either 3 or 9.
The number 52632 is divisible by 9 because the sum of its digits (18) is divisible by 9

Divisibility test for numbers by 10

Examples.
8200 is divisible by 10 and 100;
542000 is divisible by 10, 100, 1000.

3. Signs of divisibility of natural numbers by 7, 11, 12, 13, 14, 19, 25.50.

From additional literature we found confirmation of the correctness of the criteria we formulated for divisibility of natural numbers by 4, 6, 8, 15, 25, 50, 100, 1000. We also found several signs of divisibility by 7:
1) A natural number is divisible by 7 if and only if the difference between the number of thousands and the number expressed by the last three digits is divisible by 7.
Examples:
478009 is divisible by 7 because 478-9=469, 469 is divisible by 7.
479345 is not divisible by 7, because 479-345=134, 134 is not divisible by 7.
2) A natural number is divisible by 7 if the sum of the double number to the tens and the remaining number is divisible by 7.
Examples:
4592 is divisible by 7 because 45·2=90, 90+92=182, 182 is divisible by 7.
57384 is not divisible by 7, because 573·2=1146, 1146+84=1230, 1230 is not divisible by 7.
3) A three-digit natural number of the form aba will be divisible by 7 if a+b is divisible by 7.
Examples:
252 is divisible by 7 because 2+5=7, 7/7.
636 is not divisible by 7 because 6+3=9, 9 is not divisible by 7.
4) A three-digit natural number of the form baa will be divisible by 7 if the sum of the digits of the number is divisible by 7.
Examples:
455 is divisible by 7 because 4+5+5=14, 14/7.
244 is not divisible by 7, because 2+4+4=12, 12 is not divisible by 7.
5) A three-digit natural number of the form aab will be divisible by 7 if 2a-b is divisible by 7.
Examples:
882 is divisible by 7 because 8+8-2=14, 14/7.
996 is not divisible by 7, because 9+9-6=12, 12 is not divisible by 7.
6) A four-digit natural number of the form baa, where b is a two-digit number, will be divisible by 7 if b+2a is divisible by 7.
Examples:
2744 is divisible by 7 because 27+4+4=35, 35/7.
1955 is not divisible by 7, because 19+5+5=29, 29 is not divisible by 7.
7) A natural number is divisible by 7 if and only if the result of subtracting twice the last digit from that number without the last digit is divisible by 7.
Examples:
483 is divisible by 7 because 48-3·2=42, 42/7.
564 is not divisible by 7 because 56-4 2=48, 48 is not divisible by 7.
8) A natural number is divisible by 7 if and only if the sum of the products of the digits of the number by the corresponding remainders obtained by dividing the digit units by the number 7 is divisible by 7.
Examples:
10׃7=1 (ost 3)
100׃7=14 (ost 2)
1000׃7=142 (ost 6)
10000׃7=1428 (ost 4)
100000׃7=14285 (ost 5)
1000000׃7=142857 (rest 1) and the remainders are repeated again.
The number 1316 is divisible by 7 because 1 6 + 3 2 + 1 3 + 6 = 21, 21/7 (6 remainder from dividing 1000 by 7; 2 remainder from dividing 100 by 7; 3 remainder from dividing 10 by 7) .
The number 354722 is not divisible by 7, because... 3·5+5·4+4·6+7·2+2·3+2=81, 81 is not divisible by 7 (5 is the remainder of dividing 100,000 by 7; 4 is the remainder of dividing 10,000 by 7; 6-rest from dividing 1000 by 7; 2-rest from dividing 10 by 7).
Divisibility by 11.
1) A number is divisible by 11 if the difference between the sum of the digits in odd places and the sum of the digits in even places is a multiple of 11.
The difference can be a negative number or 0, but must be a multiple of 11. Numbering goes from left to right.
Example:
2135704 2+3+7+4=16, 1+5+0=6, 16-6=10, 10 is not a multiple of 11, which means this number is not divisible by 11.
1352736 1+5+7+6=19, 3+2+3=8, 19-8=11, 11 is a multiple of 11, which means this number is divisible by 11.
2) A natural number is divided from right to left into groups of 2 digits each and these groups are added. If the resulting sum is a multiple of 11, then the number being tested is a multiple of 11.
Example: Determine whether the number 12561714 is divisible by 11.
Let's divide the number into groups of two digits each: 12/56/17/14; 12+56+17+14=99, 99 is divisible by 11, which means this number is divisible by 11.
3) A three-digit natural number is divisible by 11 if the sum of the side digits of the number is equal to the digit in the middle. The answer will consist of those same side numbers.
Examples:
594 is divisible by 11 because 5+4=9, 9 is in the middle.
473 is divisible by 11 because 4+3=7, 7- in the middle.
861 is not divisible by 11 because 8+1=9, and in the middle there is 6.
Divisibility test by 12
A natural number is divisible by 12 if and only if it is divisible by 3 and 4 at the same time.
Examples:
636 is divisible by 3 and 4, which means it is divisible by 12.
587 is not divisible by 3 or 4, which means it is not divisible by 12.
27126 is divisible by 3 but not divisible by 4, which means it is not divisible by 12.
Tests for divisibility by 13
1) A natural number is divisible by 13 if the difference between the number of thousands and the number formed by the last three digits is divisible by 13.
Examples:
The number 465400 is divisible by 13 because... 465 - 400 = 65, 65 divided by 13.
The number 256184 is not divisible by 13, because... 256 - 184 = 72, 72 is not divisible by 13.
2) A natural number is divisible by 13 if and only if the result of subtracting the last digit multiplied by 9 from that number without the last digit is divisible by 13.
Examples:
988 is divisible by 13 because 98 - 9 8 = 26, 26 is divided by 13.
853 is not divisible by 13 because 85 - 3 9 = 58, 58 is not divisible by 13.
Divisibility test by 14
A natural number is divisible by 14 if and only if it is divisible by 2 and 7 at the same time.
Examples:
The number 45826 is divisible by 2 but not divisible by 7, which means it is not divisible by 14.
The number 1771 is divisible by 7 but not divisible by 2, which means it is not divisible by 14.
The number 35882 is divisible by 2 and 7, which means it is divisible by 14.
Divisibility test by 19
A natural number is divisible by 19 without a remainder if and only if the number of its tens added to twice the number of units is divisible by 19.
It should be taken into account that the number of tens in a number should be counted not by the digit in the tens place, but by the total number of whole tens in the entire number.
Examples:
1534 tens is 153, 4 2 = 8, 153 + 8 = 161, 161 is not divisible by 19, which means 1534 is not divisible by 19.
1824 182+4·2=190, 190/19, which means the number is 1824/19.
Test for divisibility by 25 and 50
Divide by 25 or 50 are those and only those numbers that end with two zeros or whose last two digits express a number divisible by 25 or 50, respectively.

The number 97300 ends with two zeros, which means it is divisible by both 25 and 50.

The number 79,450 is divisible by 25 and 50, since the number formed by the last two digits 50 is divisible by both 25 and 50.

4. Solving problems using divisibility criteria.

Seller in a store.

The buyer took from the store a package of milk worth 34.5 rubles, a box of cottage cheese worth 36 rubles, 6 cakes and 3 kilograms of sugar. When the cashier knocked out a check for 296 rubles, the buyer demanded to check the calculation and correct the error. How did the buyer determine that the invoice was incorrect?

Solution: The cost of purchased goods of each type is expressed as a number divisible by 3 (for goods of the first two types the price is a multiple of 3, and for the rest - the number of purchased goods is a multiple of 3). If each of the terms is divisible by 3, then the amount must be divisible by 3. The number 296 is not divisible by 3, therefore the calculation is incorrect.

Apples in a boxke.

The number of apples in the box is less than 200. They can be divided equally between 2,3,4,5 and 6 children. Which maximum quantity apples maybe in the box?

Solution.

LCM(2,3,4,5,6) = 60.

60s< 200, значит максимальное количество яблок в ящике = 180

Answer: 180 apples.

5. Conclusion:

While doing the work, I became acquainted with the history of the development of signs of divisibility, formulated signs of divisibility of natural numbers by 4, 6, 8, 15, 25,50 and found confirmation of this from additional literature. I also became convinced that there are other signs of divisibility of natural numbers (by 7, 11, 12, 13, 14, 19, 37), which confirmed the correctness of the hypothesis about the existence of other signs of divisibility of natural numbers.

List of used literature (sources):

1. Galkin V.A. Problems on the topic “Divisibility criteria”. // Mathematics, 1999.-№5.-P.9.

2. Gusev V.A., Orlov A.I., Rosenthal A.L. Extracurricular work in mathematics in grades 6-8. - M.: Education, 1984.

3. Kaplun L.M. GCD and LCM in problems. // Mathematics, 1999.- No. 7. - P. 4-6.

4.​ Pelman Ya.I. Mathematics is interesting! - M.: TERRA - Book club, 2006

5.​ Encyclopedic Dictionary of a Young Mathematician./ Comp. Savin A.P. - M.: Pedagogy, 1989. - P. 352.

6.​ Resources - Internet.

Divisibility of numbers. Prime and composite numbers.

One of the goals mathematics education, reflected in the federal component state standard in mathematics, is intellectual development students.

Topic: Divisibility of numbers. Prime and composite numbers" is one of those topics that, starting from grade 5, allows to a greater extent develop children's mathematical abilities. Working in a school with in-depth study of mathematics, physics and computer science, where teaching is conducted from the 7th grade, the mathematics department of our school is interested in ensuring that students in grades 5-7 become more familiar with this topic. We try to implement this in classes at the School of Young Mathematicians (SYUM), as well as in the regional summer mathematics camp, where I teach together with the teachers of our school. I tried to select tasks that would be of interest to students from grades 5 to 11. After all, the students of our school study this topic according to the program. And for the last 2 years, school graduates have been faced with problems on this topic on the Unified State Exam (in problems of type C6). Theoretical material in different cases I consider it to different extents.

Divisibility of natural numbers.

Some definitions:

A natural number a is said to be divisible by a natural number b if there is a natural number c such that a=bc. At the same time they write: a b. In this

In this case, b is called a divisor of a, and a is a multiple of b. A natural number is called prime if it has no divisors.

different from itself and from the unit (for example: 2, 3, 5, 7, etc.). A number is called composite if it is not prime. The unit is neither simple nor composite.

A number n is divisible by a prime number p if and only if p occurs among prime factors, into which n is decomposed.

The greatest common divisor of numbers a and b is called greatest number, which is both a divisor of a and a divisor of b, is denoted by GCD (a;b) or D (a;b).

The least common multiple is called smallest number, divisible by both a and b, is denoted LCM (a;b) or K (a;b).

The numbers a and b are called mutually prime, if their greatest common divisor is equal to one.

Fundamental Theorem of Arithmetic

Every natural number n can be uniquely expanded (up to the order of the factors) into a product of powers of prime factors:

n = p1 k 1 p2 k 2 pm k m

here p1, p2,…pm are various prime divisors of the number n, and k1, k2,…km are the degrees of occurrence (degrees of multiplicity) of these divisors.

Signs of divisibility

∙ A number is divisible by 2 if and only if the last digit is divisible by 2 (that is, even).

∙ A number is divisible by 3 if and only if the sum of its digits is divisible by 3.

∙ A number is divisible by 4 if and only if the two-digit number made up of the last two digits is divisible by 4.

∙ A number is divisible by 5 if and only if the last digit is divisible by 5 (that is, equal to 0 or 5).

∙ To find out whether a number is divisible by 7 (by 13), you need to divide its decimal notation from right to left into groups of 3 digits each (the most left group may contain 1 or 2 digits), then take groups with odd numbers with a minus sign, and with even numbers with a plus sign. If the resulting expression is divisible by 7 (by 13), then given number divisible by 7 (by 13).

∙ A number is divisible by 8 if and only if the three-digit number made up of the last three digits is divisible by 8.

∙ A number is divisible by 9 if and only if the sum of its digits is divisible by 9.

∙ A number is divisible by 10 if and only if the last digit is zero.

∙ A number is divisible by 11 if and only if the sum of its digits in even places in decimal notation and the sum of its digits in odd places in decimal notation give the same remainder when divided by 11.

Statements related to the divisibility of numbers.

∙ If a b and b c , then a c .

∙ If a m, then ab m.

∙ If a m and b m, then a+b m

∙ If a+.b m and a m, then b m

∙ If a m and a k, and m and k are coprime, then a mk

∙ If ab m and a are coprime to m, then b m

∙ In classes on this topic, depending on the age of the students, the place and time of the classes, I consider various tasks. I select these problems mainly from the sources that are indicated at the end of the work, including from the materials of the Perm regional tournament of young mathematicians of past years and materials II and III stages Russian Olympiad schoolchildren in mathematics from previous years.

I use the following tasks to conduct classes in grades 5, 6, 7 at SHYuM1 e when covering the topic “Divisibility of numbers. Prime and composite numbers. Signs of divisibility."

Oral tasks.

1. Add 1 digit to the left and right of the number 15 so that the number is divisible by 15.

Answer: 1155, 3150, 4155, 6150, 7155, 9150.

2. Add 1 digit to the left and right of the number 10 so that the number is divisible by 72.

Answer: 4104.

3. A certain number is divisible by 6 and 4. Does it have to be divisible by 24?

Answer: no, for example 12.

4. Find the largest natural number that is a multiple of 36 and has all its digits represented once.

Answer: 9876543120.

5. The number given is 645*7235. Replace * with a number so that the resulting number is a multiple of 3. Answer: 1, 4, 7.

6. The number 72*3* is given. Replace * with numbers so that the resulting number is a multiple of 45. Answer: 72630, 72135.

"Semi-oral" tasks.

1. How many Sundays can there be in a year?

2. In a certain month, three Sundays fell on even numbers. What day of the week was the 7th of this month?

3. Let's start counting the fingers as follows: let the first one be thumb, second - index, third - middle, fourth - ring, fifth - little finger, sixth - ring again, seventh - middle, eighth - index, ninth - thumb, tenth - index finger etc. Which finger will it be 2000?

1 SHYUM – School of Young Mathematicians – Saturday school at FMS No. 146

At what n is the number 1111...111 divisible by 7?

At what n is the number 1111...111 divisible by 999,999,999?

6. The fraction b a is reducible. Will the fraction a + − b b be reducible?

7. In the country of Anchuria, there are banknotes in circulation in denominations of 1 Anchur, 10 Anchur, 100 Anchur, 1000 Anchur. Is it possible to count 1,000,000 anchors using 500,000 banknotes?

8. Find a two-digit number whose first digit is equal to the difference between this number and a number written with the same digits, but in reverse order.

1. There can be 365 or 366 days in a year, every seventh day is Sunday, which means 365 = 52 × 7 + 1 or 366 = 52 × 7 + 2, there can be 52, or 53 if Sunday falls on the 1st day.

2. These 3 Sundays fell on the 2nd, 16th and 30th. This means that the 7th of this month will be Friday.

3. The number of fingers when counting will be repeated with a period of 8, which means that it is enough to calculate the remainder of dividing 2000 by 8. It is equal to 0. Because the index finger comes eighth, then The 2000th will be the index finger.

is an integer by 7, and 111111=7× 15873. It follows that if there are more than 6 units in the recording of a given number, then after every 6 units the next remainder is equal to 0. Thus,

a number of the form 1111...111 is divisible by 7 if and only if its quantity

digits are divisible by 6, i.e. n=7× t, where tО Z.

simultaneously. IN given number the number of units is a multiple of 9. However, the first and second such numbers 111 111 111 and 111 111 111 111 111 111 are not divisible by 999 999 999. And a number with 18 units is divisible by 999 999 999. Moreover, starting from the 18th , every 18th number is divided by 999,999,999, i.e. n=18× t, where tО N.

7. Let there be a bills in denomination of 1 anchur, b in denomination of 10 anchur, c in denomination of 100 anchur and d in denomination of 1000 anchur. We get