Let's consider a simple problem. In one farm, 846 were collected in the morning. chicken eggs. This was a common farm; it was supported by 9 families. It is necessary to divide all the eggs equally between them. How to check, without dividing, whether the number 846 is divisible by 9 without a remainder.
First, let's break down this number into digits. The number 846 is made up of 8 hundreds, 4 tens and 6 ones.
Let's start dealing with hundreds. If we put 100 eggs into nine baskets, we will have one extra egg left. That is, for every hundred eggs there will be 1 egg. Since we have 8 hundred whole eggs, there will therefore be 8 eggs left.
Now let's deal with dozens. If ten eggs are placed in nine baskets, then there will also be one extra egg left for every ten. Since there are 4 tens in our number, there will therefore be 4 eggs left.
There is no way we can put the 6 eggs that were in the one category into nine baskets, so they will also remain.
Now let's add all the eggs we have left. 8 from hundreds, 4 from tens and 6 from ones, for a total of 8+4+6=18 eggs. 18 eggs can be divided into nine baskets and there will not be a single extra egg left. Therefore, 846 eggs can be divided equally into nine baskets. This means that the number 846 is divisible by 9 without a remainder.
Divisibility test by 9
Now, we can formulate a test for the divisibility of a number by 9.
- If the sum of the digits of a number is divisible by 9 without a remainder, then the number itself is divisible by 9. If the sum of the digits of a number is not divisible by 9 without a remainder, then the number itself will not be divisible by 9 without a remainder.
Here are some examples:
The number 76,005 will be divisible by 9 without a remainder, since the sum of its constituent digits: 7+6+0+0+5=18 is divisible by 9 without a remainder.
The number 51,734 is not divisible by 9 without a remainder, since the sum of its constituent digits: 5+1+7+3+4=20 is not divisible by 9 without a remainder.
Test for divisibility by 3
In a similar way, we obtain a sign that a number is divisible by 3.
Dividing a hundred by 3 will leave one. Dividing a ten by 3 will also leave a unit. We get a copy of the situation with nine.
- If the sum of the digits of a number is divisible by 3 without a remainder, then the number itself is divisible by 3. If the sum of the digits of a number is not divisible by 3 without a remainder, then the number itself will not be divisible by 3 without a remainder.
The number 76,005 will be divisible by 3 without a remainder, since the sum of its constituent digits: 7+6+0+0+5=18 is divisible by 3 without a remainder.
The number 51,734 is not divisible by 3 without a remainder, since the sum of its constituent digits: 5+1+7+3+4=20 is not divisible by 3 without a remainder.
Let's begin to consider the topic “Divisibility test by 3”. Let's start with the formulation of the sign and give the proof of the theorem. Then we will consider the main approaches to establishing divisibility by 3 of numbers whose value is given by some expression. The section provides an analysis of the solution to the main types of problems based on the use of the test of divisibility by 3.
Test for divisibility by 3, examples
The test of divisibility by 3 is formulated simply: an integer will be divisible by 3 without a remainder if the sum of its digits is divisible by 3. If the total value of all the digits that make up a whole number is not divisible by 3, then the original number itself is not divisible by 3. You can get the sum of all the digits in an integer using addition. natural numbers.
Now let's look at examples of using the test of divisibility by 3.
Example 1
Is the number 42 divisible by 3?
Solution
In order to answer this question, we add up all the numbers that make up the number - 42: 4 + 2 = 6.
Answer: According to the divisibility test, since the sum of the digits included in the original number is divisible by three, then the original number itself is divisible by 3.
In order to answer the question of whether the number 0 is divisible by 3, we need the divisibility property, according to which zero is divisible by any integer. It turns out that zero is divisible by three.
There are problems for which it is necessary to use the test of divisibility by 3 several times.
Example 2
Show that the number 907 444 812 divisible by 3.
Solution
Let's find the sum of all the digits that form the original number: 9 + 0 + 7 + 4 + 4 + 4 + 8 + 1 + 2 = 39 . Now we need to determine whether the number 39 is divisible by 3. Once again we add up the numbers that make up this number: 3 + 9 = 12 . We just have to add the numbers again to get the final answer: 1 + 2 = 3 . The number 3 is divisible by 3
Answer: original number 907 444 812 is also divisible by 3.
Example 3
Is the number divisible by 3? − 543 205 ?
Solution
Let's calculate the sum of the digits that make up the original number: 5 + 4 + 3 + 2 + 0 + 5 = 19 .
Now let's calculate the sum of the digits of the resulting number: 1 + 9 = 10 .
In order to get the final answer, we find the result of one more addition: 1 + 0 = 1
.
Answer: 1 is not divisible by 3, which means the original number is not divisible by 3.
In order to determine whether a given number is divisible by 3 without a remainder, we can divide the given number by 3. If you divide the number − 543 205 from the example discussed above with a column of three, then we will not get an integer in the answer. This also means that − 543 205 cannot be divided by 3 without a remainder.
Proof of the test of divisibility by 3
Here we will need the following skills: decomposition of a number into digits and the rule of multiplication by 10, 100, etc. In order to carry out the proof, we need to obtain a representation of the number a of the form , Where a n , a n − 1 , … , a 0- these are the numbers that are located from left to right in the notation of a number.
Here's an example using a specific number: 528 = 500 + 20 + 8 = 5 100 + 2 10 + 8.
Let's write down a series of equalities: 10 = 9 + 1 = 3 3 + 1, 100 = 99 + 1 = 33 3 + 1, 1,000 = 999 + 1 = 333 3 + 1, and so on.
Now let’s substitute these equalities instead of 10, 100 and 1000 into the equalities given earlier a = a n 10 n + a n - 1 10 n - 1 + … + a 2 10 2 + a 1 10 + a 0.
This is how we arrived at equality:
a = a n 10 n + … + a 2 100 + a 1 10 + a 0 = = a n 33. . . . 3 3 + 1 + … + a 2 33 3 + 1 + a 1 3 3 + 1 + a 0
Now let’s apply the properties of addition and the properties of multiplication of natural numbers in order to rewrite the resulting equality as follows:
a = a n · 33 . . . 3 · 3 + 1 + . . . + + a 2 · 33 · 3 + 1 + a 1 · 3 · 3 + 1 + a 0 = = 3 · 33 . . . 3 a n + a n + . . . + + 3 · 33 · a 2 + a 2 + 3 · 3 · a 1 + a 1 + a 0 = = 3 · 33 . . . 3 a n + . . . + + 3 · 33 · a 2 + 3 · 3 · a 1 + + a n + . . . + a 2 + a 1 + a 0 = = 3 33 . . . 3 · a n + … + 33 · a 2 + 3 · a 1 + + a n + . . . + a 2 + a 1 + a 0
Expression a n + . . . + a 2 + a 1 + a 0 is the sum of the digits of the original number a. Let us introduce a new short notation for it A. We get: A = a n + . . . + a 2 + a 1 + a 0 .
In this case, the representation of the number is a = 3 33. . . 3 a n + . . . + 33 · a 2 + 3 · a 1 + A takes the form that will be convenient for us to use to prove the test for divisibility by 3.
Definition 1
Now recall the following properties of divisibility:
- necessary and sufficient condition for an integer a to be divisible by an integer
b , is the condition by which the modulus of the number a is divided by the modulus of the number b; - if in equality a = s + t all terms except one are divisible by some integer b, then this one term is also divisible by b.
We have laid the groundwork for proving the divisibility test by 3. Now let’s formulate this feature in the form of a theorem and prove it.
Theorem 1
In order to assert that the integer a is divisible by 3, it is necessary and sufficient for us that the sum of the digits that forms the notation of the number a is divisible by 3.
Evidence 1
If we take the value a = 0, then the theorem is obvious.
If we take a number a that is different from zero, then the modulus of the number a will be a natural number. This allows us to write the following equality:
a = 3 · 33 . . . 3 a n + . . . + 33 · a 2 + 3 · a 1 + A , where A = a n + . . . + a 2 + a 1 + a 0 - the sum of the digits of the number a.
Since the sum and product of integers is an integer, then
33. . . 3 a n + . . . + 33 · a 2 + 3 · a 1 is an integer, then by definition of divisibility the product is 3 · 33. . . 3 a n + . . . + 33 · a 2 + 3 · a 1 is divisible by 3
for any a 0 , a 1 , … , a n.
If the sum of the digits of a number a divided by 3 , that is, A divided by 3 , then due to the divisibility property indicated before the theorem, a is divided by 3 , hence, a divided by 3 . So the sufficiency is proven.
If a divided by 3
, then a is also divisible by 3
, then due to the same property of divisibility, the number
A divided by 3
, that is, the sum of the digits of a number a divided by 3
. The necessity has been proven.
Other cases of divisibility by 3
Integers can be specified as the value of some expression that contains a variable, given a certain value of that variable. Thus, for some natural number n, the value of the expression 4 n + 3 n - 1 is a natural number. In this case, direct division by 3 cannot give us an answer to the question whether a number is divisible by 3 . Application of the divisibility test for 3 may also be difficult. Let's look at examples of such problems and look at methods for solving them.
Several approaches can be used to solve such problems. The essence of one of them is as follows:
- we represent the original expression as a product of several factors;
- find out whether at least one of the factors can be divided by 3 ;
- Based on the divisibility property, we conclude that the entire product is divisible by 3 .
When solving, one often has to resort to using Newton's binomial formula.
Example 4
Is the value of the expression 4 n + 3 n - 1 divisible by 3 under any natural n?
Solution
Let us write the equality 4 n + 3 n - 4 = (3 + 1) n + 3 n - 4 . Let's apply Newton's binomial formula:
4 n + 3 n - 4 = (3 + 1) n + 3 n - 4 = = (C n 0 3 n + C n 1 3 n - 1 1 + . . . + + C n n - 2 3 2 · 1 n - 2 + C n n - 1 · 3 · 1 n - 1 + C n n · 1 n) + + 3 n - 4 = = 3 n + C n 1 · 3 n - 1 · 1 + . . . + C n n - 2 · 3 2 + n · 3 + 1 + + 3 n - 4 = = 3 n + C n 1 · 3 n - 1 · 1 + . . . + C n n - 2 3 2 + 6 n - 3
Now let's take it out 3 outside the brackets: 3 · 3 n - 1 + C n 1 · 3 n - 2 + . . . + C n n - 2 · 3 + 2 n - 1 . The resulting product contains the multiplier 3 , and the value of the expression in parentheses for natural n represents a natural number. This allows us to assert that the resulting product and the original expression 4 n + 3 n - 1 is divided by 3 .
Answer: Yes.
We can also use the method of mathematical induction.
Example 5
Prove using the method of mathematical induction that for any natural number
n the value of the expression n n 2 + 5 is divided by 3
.
Solution
Let's find the value of the expression n · n 2 + 5 when n=1: 1 · 1 2 + 5 = 6 . 6 is divisible by 3 .
Now suppose that the value of the expression n n 2 + 5 at n = k divided by 3 . In fact, we will have to work with the expression k k 2 + 5, which we expect to be divisible by 3 .
Considering that k k 2 + 5 is divisible by 3 , we will show that the value of the expression n · n 2 + 5 at n = k + 1 divided by 3 , that is, we will show that k + 1 k + 1 2 + 5 is divisible by 3 .
Let's perform the transformations:
k + 1 k + 1 2 + 5 = = (k + 1) (k 2 + 2 k + 6) = = k (k 2 + 2 k + 6) + k 2 + 2 k + 6 = = k (k 2 + 5 + 2 k + 1) + k 2 + 2 k + 6 = = k (k 2 + 5) + k 2 k + 1 + k 2 + 2 k + 6 = = k (k 2 + 5) + 3 k 2 + 3 k + 6 = = k (k 2 + 5) + 3 k 2 + k + 2
The expression k · (k 2 + 5) is divided by 3 and the expression 3 k 2 + k + 2 is divided by 3 , so their sum is divided by 3 .
So we proved that the value of the expression n · (n 2 + 5) is divisible by 3 for any natural number n.
Now let's look at the approach to proving divisibility by 3 , which is based on the following algorithm of actions:
- we show that the value of this expression with variable n for n = 3 m, n = 3 m + 1 and n = 3 m + 2, Where m– an arbitrary integer, divisible by 3 ;
- we conclude that the expression will be divisible by 3 for any integer n.
In order not to distract attention from minor details, we will apply this algorithm to the solution of the previous example.
Example 6
Show that n · (n 2 + 5) is divisible by 3 for any natural number n.
Solution
Let's pretend that n = 3 m. Then: n · n 2 + 5 = 3 m · 3 m 2 + 5 = 3 m · 9 m 2 + 5. The product we received contains a multiplier 3 , therefore the product itself is divided into 3 .
Let's pretend that n = 3 m + 1. Then:
n · n 2 + 5 = 3 m · 3 m 2 + 5 = (3 m + 1) · 9 m 2 + 6 m + 6 = = 3 m + 1 · 3 · (2 m 2 + 2 m + 2)
The product we received is divided into 3 .
Let's assume that n = 3 m + 2. Then:
n · n 2 + 5 = 3 m + 1 · 3 m + 2 2 + 5 = 3 m + 2 · 9 m 2 + 12 m + 9 = = 3 m + 2 · 3 · 3 m 2 + 4 m + 3
This work is also divided into 3 .
Answer: So we proved that the expression n n 2 + 5 is divisible by 3 for any natural number n.
Example 7
Is it divisible by 3 the value of the expression 10 3 n + 10 2 n + 1 for some natural number n.
Solution
Let's pretend that n=1. We get:
10 3 n + 10 2 n + 1 = 10 3 + 10 2 + 1 = 1000 + 100 + 1 = 1104
Let's pretend that n=2. We get:
10 3 n + 10 2 n + 1 = 10 6 + 10 4 + 1 = 1000 000 + 10000 + 1 = 1010001
So we can conclude that for any natural n we will get numbers that are divisible by 3. This means that 10 3 n + 10 2 n + 1 for any natural number n is divisible by 3.
Answer: Yes
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Signs of divisibility of numbers it is useful to know 2, 3, 4, 5, 6, 8, 9, 10, 11, 25 and other numbers to quickly solve problems on digital notation of numbers. Instead of dividing one number by another, it is enough to check a number of signs on the basis of which you can unambiguously determine whether one number is divisible by another (whether it is a multiple) or not.
Basic signs of divisibility
Let's give basic signs of divisibility of numbers:
- Divisibility test for a number by “2” A number is divisible by 2 if the number is even (the last digit is 0, 2, 4, 6 or 8)
Example: The number 1256 is a multiple of 2 because it ends in 6. But the number 49603 is not evenly divisible by 2 because it ends in 3. - Divisibility test for a number by “3” A number is divisible by 3 if the sum of its digits is divisible by 3
Example: The number 4761 is divisible by 3, since the sum of its digits is 18 and it is divisible by 3. And the number 143 is not a multiple of 3, since the sum of its digits is 8 and it is not divisible by 3. - Divisibility test for a number by “4” A number is divisible by 4 if the last two digits of the number are zero or the number made up of the last two digits is divisible by 4
Example: The number 2344 is a multiple of 4, since 44 / 4 = 11. And the number 3951 is not divisible by 4, since 51 is not divisible by 4. - Divisibility test for a number by “5” A number is divisible by 5 if the last digit of the number is 0 or 5
Example: The number 5830 is divisible by 5 because it ends in 0. But the number 4921 is not divisible by 5 because it ends in 1. - Divisibility test for a number by “6” A number is divisible by 6 if it is divisible by 2 and 3.
Example: The number 3504 is a multiple of 6 because it ends in 4 (divisible by 2) and the sum of the digits of the number is 12 and it is divisible by 3 (divisible by 3). And the number 5432 is not completely divisible by 6, although the number ends in 2 (the criterion of divisibility by 2 is observed), however, the sum of the digits is equal to 14 and it is not completely divisible by 3. - Divisibility test for a number by “8” A number is divisible by 8 if the last three digits of the number are zero or the number made up of the last three digits of the number is divisible by 8
Example: The number 93112 is divisible by 8, since the number 112 / 8 = 14. And the number 9212 is not a multiple of 8, since 212 is not divisible by 8. - Divisibility test for a number by “9” A number is divisible by 9 if the sum of its digits is divisible by 9
Example: The number 2916 is a multiple of 9, since the sum of the digits is 18 and it is divisible by 9. And the number 831 is not divisible by 9, since the sum of the digits of the number is 12 and it is not divisible by 9. - Test for divisibility of a number by “10” A number is divisible by 10 if it ends in 0
Example: The number 39590 is divisible by 10 because it ends in 0. And the number 5964 is not divisible by 10 because it does not end in 0. - Test for the divisibility of a number by “11” A number is divisible by 11 if the sum of the digits in odd places is equal to the sum of the digits in even places or the sums must differ by 11
Example: The number 3762 is divisible by 11, since 3 + 6 = 7 + 2 = 9. But the number 2374 is not divisible by 11, since 2 + 7 = 9, and 3 + 4 = 7. - Divisibility test for a number by “25” A number is divisible by 25 if it ends in 00, 25, 50 or 75
Example: The number 4950 is a multiple of 25 because it ends in 50. And 4935 is not divisible by 25 because it ends in 35.
Signs of divisibility by a composite number
To find out if it is dividing given number into a composite number, you need to factor this composite number into coprime factors, the signs of divisibility of which are known. Mutually prime numbers- these are numbers that have no common divisors other than 1. For example, a number is divisible by 15 if it is divisible by 3 and 5.
Consider another example of a compound divisor: a number is divisible by 18 if it is divisible by 2 and 9. In in this case you cannot expand 18 into 3 and 6, since they are not relatively prime, since they have a common divisor 3. Let's see this with an example.
The number 456 is divisible by 3, since the sum of its digits is 15, and divisible by 6, since it is divisible by both 3 and 2. But if you divide 456 by 18 manually, you get a remainder. If you check the signs of divisibility by 2 and 9 for the number 456, you can immediately see that it is divisible by 2, but not divisible by 9, since the sum of the digits of the number is 15 and it is not divisible by 9.
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In mathematics lessons, when studying the topic “Signs of divisibility”, where we became acquainted with the signs of divisibility by 2; 5; 3; 9; 10, I was interested in whether there are signs of divisibility by other numbers, and whether there is a universal method of divisibility by any natural number. Therefore, I began research work on this topic.
Purpose of the study: study of signs of divisibility of natural numbers up to 100, addition of already known signs of divisibility of natural numbers by whole, studied at school.
To achieve the goal, we set tasks:
Collect, study and systematize material about the signs of divisibility of natural numbers, using various sources of information.
Find a universal test for divisibility by any natural number.
Learn to use Pascal's divisibility test to determine the divisibility of numbers, and also try to formulate tests for divisibility by any natural number.
Object of study: divisibility of natural numbers.
Subject of study: signs of divisibility of natural numbers.
Research methods: collection of information; working with printed materials; analysis; synthesis; analogy; survey; survey; systematization and generalization of material.
Research hypothesis: If it is possible to determine the divisibility of natural numbers by 2, 3, 5, 9, 10, then there must be signs by which one can determine the divisibility of natural numbers by other numbers.
Novelty carried out research work thing is this work systematizes knowledge about divisibility criteria and the universal method of divisibility of natural numbers.
Practical significance: the material of this research work can be used in grades 6 - 8 in elective classes when studying the topic “Divisibility of Numbers”.
Chapter I. Definition and properties of divisibility of numbers
1.1.Definitions of the concepts of divisibility and signs of divisibility, properties of divisibility.
Number theory is a branch of mathematics that studies the properties of numbers. The main object of number theory is natural numbers. Their main property, which is considered by number theory, is divisibility. Definition: An integer a is divisible by an integer b that is not equal to zero if there is an integer k such that a = bk (for example, 56 is divisible by 8, since 56 = 8x7). Divisibility test- a rule that allows you to determine whether a given natural number is divisible by some other numbers by an integer, i.e. without a trace.
Divisibility properties:
Any number a other than zero is divisible by itself.
Zero is divisible by any b not equal to zero.
If a is divisible by b (b0) and b is divisible by c (c0), then a is divisible by c.
If a is divisible by b (b0) and b is divisible by a (a0), then a and b are either equal or opposite numbers.
1.2. Properties of divisibility of a sum and a product:
If in a sum of integers each term is divisible by a certain number, then the sum is divided by that number.
2) If in the difference of integers the minuend and the subtrahend are divisible by a certain number, then the difference is also divisible by a certain number.
3) If in the sum of integers all terms except one are divisible by a certain number, then the sum is not divisible by this number.
4) If in a product of integers one of the factors is divisible by a certain number, then the product is also divisible by this number.
5) If in a product of integers one of the factors is divisible by m and the other by n, then the product is divisible by mn.
In addition, while studying the signs of divisibility of numbers, I became acquainted with the concept "digital root number". Let's take a natural number. Let's find the sum of its digits. We also find the sum of the digits in the result, and so on until we get a single-digit number. The resulting result is called the digital root of the number. For example, the digital root of the number 654321 is 3: 6+5+4+3+2+1=21.2+1=3. And now you can think about the question: “What signs of divisibility exist and is there a universal sign of the divisibility of one number by another?”
Chapter II. Divisibility criteria for natural numbers.
2.1. Signs of divisibility by 2,3,5,9,10.
Among the signs of divisibility, the most convenient and well-known of school course 6th grade mathematics:
Divisibility by 2. If a natural number ends in an even digit or zero, then the number is divisible by 2. The number 52738 is divisible by 2, since the last digit is 8.
Divisibility by 3 . If the sum of the digits of a number is divisible by 3, then the number is divisible by 3 (the number 567 is divisible by 3, since 5+6+7 = 18, and 18 is divisible by 3.)
Divisibility by 5. If a natural number ends in 5 or zero, then the number is divisible by 5 (the number 130 and 275 are divisible by 5, since the last digits of the numbers are 0 and 5, but the number 302 is not divisible by 5, since the last digit the numbers are not 0 and 5).
Divisible by 9. If the sum of the digits is divisible by 9, then the number is divisible by 9 (676332 is divisible by 9 because 6+7+6+3+3+2=27, and 27 is divisible by 9).
Divisibility by 10 . If a natural number ends in 0, then this number is divisible by 10 (230 is divisible by 10, since the last digit of the number is 0).
2.2. Signs of divisibility by 4,6,8,11,12,13, etc.
After working with various sources, I learned other signs of divisibility. I will describe some of them.
Division by 6 . We need to check the divisibility of the number we are interested in by 2 and 3. A number is divisible by 6 if and only if it is even and its digital root is divisible by 3. (For example, 678 is divisible by 6, since it is even and 6 +7+8=21, 2+1=3) Another sign of divisibility: a number is divisible by 6 if and only if the quadruple number of tens added to the number of units is divisible by 6. (73.7*4+3=31, 31 is not divisible by 6, which means 7 is not divisible by 6.)
Division by 8. A number is divisible by 8 if and only if its last three digits form a number divisible by 8. (12,224 is divisible by 8 because 224:8=28). A three-digit number is divisible by 8 if and only if the number of units added to twice the number of tens and quadruple the number of hundreds is divisible by 8. For example, 952 is divisible by 8 since 9 * 4 + 5 * 2 + 2 = 48 is divisible by 8 .
Division by 4 and 25. If the last two digits are zeros or express a number divisible by 4 and/or 25, then the number is divisible by 4 and/or 25 (the number 1500 is divisible by 4 and 25, since it ends with two zeros, the number 348 is divisible by 4, since 48 is divisible by 4, but this number is not divisible by 25, because 48 is not divisible by 25, the number 675 is divisible by 25, because 75 is divisible by 25, but not divisible by 4, i.e. .k. 75 is not divisible by 4).
Knowing the basic signs of divisibility by prime numbers, you can derive the signs of divisibility by composite numbers:
Divisibility test for11 . If the difference between the sum of digits in even places and the sum of digits in odd places is divisible by 11, then the number is divisible by 11 (the number 593868 is divisible by 11, since 9 + 8 + 8 = 25, and 5 + 3 + 6 = 14, their difference is 11, and 11 is divided by 11).
Test for divisibility by 12: a number is divisible by 12 if and only if the last two digits are divisible by 4 and the sum of the digits is divisible by 3.
because 12= 4 ∙ 3, i.e. the number must be divisible by 4 and 3.
Test for divisibility by 13: A number is divisible by 13 if and only if the alternating sum of numbers formed by successive triplets of digits of the given number is divisible by 13. How do you know, for example, that the number 354862625 is divisible by 13? 625-862+354=117 is divisible by 13, 117:13=9, which means the number 354862625 is divisible by 13.
Test for divisibility by 14: A number is divisible by 14 if and only if it ends in an even digit and when the result of subtracting twice the last digit from that number without the last digit is divisible by 7.
because 14= 2 ∙ 7, i.e. the number must be divisible by 2 and 7.
Test for divisibility by 15: A number is divisible by 15 if and only if it ends in 5 and 0 and the sum of the digits is divisible by 3.
because 15= 3 ∙ 5, i.e. the number must be divisible by 3 and 5.
Test for divisibility by 18: A number is divisible by 18 if and only if it ends in an even digit and the sum of its digits is divisible by 9.
because18= 2 ∙ 9, i.e. the number must be divisible by 2 and 9.
Test for divisibility by 20: A number is divisible by 20 if and only if the number ends in 0 and the penultimate digit is even.
because 20 = 10 ∙ 2 i.e. the number must be divisible by 2 and 10.
Test for divisibility by 25: a number containing at least three digits is divisible by 25 if and only if the number formed by the last two digits is divisible by 25.
Divisibility test for30 .
Divisibility test for59 . A number is divisible by 59 if and only if the number of tens added to the number of units multiplied by 6 is divisible by 59. For example, 767 is divisible by 59, since 76 + 6*7 = 118 and 11 + 6* are divisible by 59 8 = 59.
Divisibility test for79 . A number is divisible by 79 if and only if the number of tens added to the number of units multiplied by 8 is divisible by 79. For example, 711 is divisible by 79, since 79 is divisible by 71 + 8*1 = 79.
Divisibility test for99. A number is divisible by 99 if and only if the sum of numbers that form groups of two digits (starting with ones) is divisible by 99. For example, 12573 is divisible by 99, since 1 + 25 + 73 = 99 is divisible by 99.
Divisibility test for100 . Only those numbers whose last two digits are zeros are divisible by 100.
Divisibility test by 125: a number containing at least four digits is divisible by 125 if and only if the number formed by the last three digits is divisible by 125.
All of the above characteristics are summarized in table form. (Annex 1)
2.3 Tests for divisibility by 7.
1) Let’s take the number 5236 for testing. Let’s write this number as follows: 5236=5*1000+2*100+3*10+6=10 3 *5+10 2 *2+10*3+6 (“systematic » form of writing a number), and everywhere we replace the base 10 with the base 3); 3 3 *5 + 3 2 *2 + 3*3 + 6 = 168. If the resulting number is divisible (not divisible) by 7, then this number is also divisible (not divisible) by 7. Since 168 is divisible by 7, then 5236 is divisible by 7. 68:7=24, 5236:7=748.
2) In this sign you need to act exactly the same as in the previous one, with the only difference that the multiplication should start from the far right and multiply not by 3, but by 5. (5236 is divisible by 7, since 6 * 5 3 +3*5 2 +2*5+5=840, 840:7=120)
3) This sign is less easy to implement in the mind, but is also very interesting. Double the last digit and subtract the second from the right, double the result and add the third from the right, etc., alternating subtraction and addition and decreasing each result, where possible, by 7 or a multiple of seven. If the final result is divisible (not divisible) by 7, then the tested number is divisible (not divisible) by 7. ((6*2-3) *2+2) *2-5=35, 35:7=5.
4) A number is divisible by 7 if and only if the alternating sum of numbers formed by successive triplets of digits of a given number is divisible by 7. How do you know, for example, that the number 363862625 is divisible by 7? 625-862+363=126 is divisible by 7, 126:7=18, which means the number 363862625 is divisible by 7, 363862625:7=51980375.
5) One of the oldest signs of divisibility by 7 is as follows. The digits of the number must be taken in reverse order, from right to left, multiplying the first digit by 1, the second by 3, the third by 2, the fourth by -1, the fifth by -3, the sixth by -2, etc. (if the number of characters is more than 6, the sequence of factors 1, 3, 2, -1, -3, -2 should be repeated as many times as necessary). The resulting products must be added up. The original number is divisible by 7 if the calculated sum is divisible by 7. Here, for example, is what this sign gives for the number 5236. 1*6+3*3+2*2+5*(-1) =14. 14: 7=2, which means the number 5236 is divisible by 7.
6) A number is divisible by 7 if and only if triple the number of tens added to the number of units is divisible by 7. For example, 154 is divisible by 7, since the number 49 is 7, which we obtain from this criterion: 15* 3 + 4 = 49.
2.4.Pascal's test.
A great contribution to the study of signs of divisibility of numbers was made by B. Pascal (1623-1662), a French mathematician and physicist. He found an algorithm for finding signs of divisibility of any integer by any other integer, which he published in the treatise “On the nature of the divisibility of numbers.” Almost all currently known divisibility tests are a special case of Pascal’s test: “If the sum of the remainders when dividing a numbera by digits per numberV divided byV , then the numberA divided byV ». Knowing him is useful even today. How can we prove the divisibility tests formulated above (for example, the familiar test of divisibility by 7)? I'll try to answer this question. But first, let’s agree on a way to write numbers. To write down a number whose digits are indicated by letters, we agree to draw a line over these letters. Thus, abcdef will denote a number having f units, e tens, d hundreds, etc.:
abcdef = a . 10 5 + b. 10 4 + c. 10 3 + d. 10 2 + e. 10 + f. Now I will prove the test for divisibility by 7 formulated above. We have:
10 9 10 8 10 7 10 6 10 5 10 4 10 3 10 2 10 1
1 2 3 1 -2 -3 -1 2 3 1
(remainders from division by 7).
As a result, we get the 5th rule formulated above: to find out the remainder of dividing a natural number by 7, you need to sign the coefficients (division remainders) under the digits of this number from right to left: then you need to multiply each digit by the coefficient below it and add the resulting products; the sum found will have the same remainder when divided by 7 as the number taken.
Let's take the numbers 4591 and 4907 as an example and, acting as indicated in the rule, we will find the result:
-1 2 3 1
4+10+27+1 = 38 - 4 = 34: 7 = 4 (remainder 6) (not divisible by 7)
-1 2 3 1
4+18+0+7 = 25 - 4 = 21: 7 = 3 (divisible by 7)
In this way you can find a test for divisibility by any number T. You just need to find which coefficients (division remainders) should be signed under the digits of the taken number A. To do this, you need to replace each power of ten by 10, if possible, with the same remainder when divided by T, same as the number 10. When T= 3 or t = 9, these coefficients turned out to be very simple: they are all equal to 1. Therefore, the test for divisibility by 3 or 9 turned out to be very simple. At T= 11, the coefficients were also not complicated: they are alternately equal to 1 and - 1. And when t =7 the coefficients turned out to be more complicated; Therefore, the test for divisibility by 7 turned out to be more complex. Having examined the signs of division up to 100, I was convinced that the most complex coefficients for natural numbers are 23 (from 10 23 the coefficients are repeated), 43 (from 10 39 the coefficients are repeated).
All of the listed signs of divisibility of natural numbers can be divided into 4 groups:
1 group- when the divisibility of numbers is determined by the last digit(s) - these are signs of divisibility by 2, by 5, by a digit unit, by 4, by 8, by 25, by 50.
2nd group- when the divisibility of numbers is determined by the sum of the digits of the number - these are signs of divisibility by 3, by 9, by 7, by 37, by 11 (1 sign).
3 group- when the divisibility of numbers is determined after performing some actions on the digits of the number - these are signs of divisibility by 7, by 11 (1 sign), by 13, by 19.
4 group- when other signs of divisibility are used to determine the divisibility of a number - these are signs of divisibility by 6, by 15, by 12, by 14.
experimental part
Survey
The survey was conducted among students in grades 6 and 7. 58 students of the municipal educational institution Karaidel secondary school No. 1 of the MR Karaidel district of the Republic of Belarus took part in the survey. They were asked to answer the following questions:
Do you think there are other signs of divisibility different from those studied in class?
Are there any signs of divisibility for other natural numbers?
Would you like to know these signs of divisibility?
Do you know any signs of divisibility of natural numbers?
The results of the survey showed that 77% of respondents believe that there are other signs of divisibility besides those studied at school; 9% do not think so, 13% of respondents found it difficult to answer. To the second question, “Would you like to know the divisibility tests for other natural numbers?” 33% answered affirmatively, 17% of respondents answered “No” and 50% found it difficult to answer. To the third question, 100% of respondents answered in the affirmative. The fourth question was answered positively by 89%, and “No” was answered by 11% of students who participated in the survey during the research work.
Conclusion
Thus, during the work the following tasks were solved:
studied theoretical material on this issue;
in addition to the signs known to me for 2, 3, 5, 9 and 10, I learned that there are also signs of divisibility by 4, 6, 7, 8, 11, 12, 13, 14, 15, 19, etc.;
3) Pascal’s test was studied - a universal test of divisibility by any natural number;
Working with different sources, analyzing the material found on the topic under study, I became convinced that there are signs of divisibility by other natural numbers. For example, on 7, 11, 12, 13, 14, 19, 37, which confirmed the correctness of my hypothesis about the existence of other signs of the divisibility of natural numbers. I also found out that there is a universal criterion for divisibility, the algorithm of which was found by the French mathematician Pascal Blaise and published it in his treatise “On the nature of the divisibility of numbers.” Using this algorithm, you can obtain a test for divisibility by any natural number.
The result of research work became a systematized material in the form of a table “Signs of divisibility of numbers”, which can be used in mathematics lessons, in extracurricular activities in order to prepare students for solving Olympiad problems, in preparing students for the Unified State Exam and Unified State Exam.
In the future, I plan to continue working on the application of divisibility tests for numbers to solving problems.
List of sources used
Vilenkin N.Ya., Zhokhov V.I., Chesnokov A.S., Shvartsburd S.I. Mathematics. 6th grade: educational. for general education institutions /— 25th ed., erased. - M.: Mnemosyne, 2009. - 288 p.
Vorobiev V.N. Signs of divisibility.-M.: Nauka, 1988.-96 p.
Vygodsky M.Ya. Guide to elementary mathematics. - Elista.: Dzhangar, 1995. - 416 p.
Gardner M. Mathematical leisure. / Under. Ed. Y.A. Smorodinsky. - M.: Onyx, 1995. - 496 p.
Gelfman E.G., Beck E.F. etc. The case of divisibility and other stories: Tutorial in mathematics for 6th grade. - Tomsk: Tomsk University Publishing House, 1992. - 176 p.
Gusev V. A., Mordkovich A. G. Mathematics: Reference. materials: Book. for students. - 2nd ed. - M.: Education, 1990. - 416 p.
Gusev V.A., Orlov A.I., Rosenthal A.V. Extracurricular work in mathematics in grades 6-8. Moscow: Education, 1984. - 289 p.
Depman I.Ya., Vilenkin N.Ya. Behind the pages of a mathematics textbook. M.: Education, 1989. - 97 p.
Kulanin E.D. Mathematics. Directory. -M.: EKSMO-Press, 1999-224 p.
Perelman Ya.I. Entertaining algebra. M.: Triada-Litera, 1994. -199s.
Tarasov B.N. Pascal. -M.: Mol. Guard, 1982.-334 p.
http://dic.academic.ru/ (Wikipedia - the free encyclopedia).
http://www.bymath.net (encyclopedia).
Annex 1
TABLE OF SIGNIFICANCE SIGNS
|
Sign |
Example |
|
|
The number ends with an even digit. |
………………2(4,6,8,0) |
|
|
The sum of the numbers is divisible by 3. |
3+7+8+0+1+5 = 24. 24:3 |
|
|
A number whose last two digits are zeros or divisible by 4. |
………………12 |
|
|
The number ends with the number 5 or 0. |
………………0(5) |
|
|
The number ends with an even digit and the sum of the digits is divisible by 3. |
375018: 8-even number 3+7+5+0+1+8 = 24. 24:3 |
|
|
The result of subtracting twice the last digit from that number without the last digit is divided by 7. |
36 - (2 × 4) = 28, 28:7 |
|
|
Its last three digits are zeros or form a number that is divisible by 8. |
……………..064 |
|
|
The sum of its digits is divisible by 9. |
3+7+8+0+1+5+3=27. 27:9 |
|
|
Number ends in zero |
………………..0 |
|
|
The sum of the digits of a number with alternating signs is divisible by 11. |
1 — 8 + 2 — 9 + 1 — 9 = −22 |
|
|
The last two digits of the number are divisible by 4 and the sum of the digits is divisible by 3. |
2+1+6=9, 9:3 and 16:4 |
|
|
The number of tens of a given number added to four times the number of units is a multiple of 13. |
84 + (4 × 5) = 104, |
|
|
A number ends with an even digit and when the result of subtracting twice the last digit from that number without the last digit is divisible by 7. |
364: 4 - even number 36 - (2 × 4) = 28, 28:7 |
|
|
The number 5 is divided by 0 and the sum of the digits is divisible by 3. |
6+3+4+8+0=21, 21:3 |
|
|
Its last four digits are zeros or form a number that is divisible by 16. |
…………..0032 |
|
|
The number of tens of a given number added to the number of units increased by 12 times is a multiple of 17. |
29053→2905+36=2941→294+12= 306→30+72=102→10+24=34. Since 34 is divisible by 17, then 29053 is divisible by 17 |
|
|
The number ends with an even digit and the sum of its digits is divisible by 9. |
2034: 4 - even number |
|
|
The number of tens of a given number added to twice the number of units is a multiple of 19 |
64 + (6 × 2) = 76, |
|
|
The number ends in 0 and the penultimate digit is even |
…………………40 |
|
|
A number consisting of the last two digits is divisible by 25 |
…………….75 |
|
|
A number is divisible by 30 if and only if it ends in 0 and the sum of all digits is divisible by 3. |
……………..360 |
|
|
A number is divisible by 59 if and only if the number of tens added to the number of units multiplied by 6 is divisible by 59. |
For example, 767 is divisible by 59, since 76 + 6*7 = 118 and 11 + 6*8 = 59 are divisible by 59. |
|
|
A number is divisible by 79 if and only if the number of tens added to the number of units multiplied by 8 is divisible by 79. |
For example, 711 is divisible by 79, since 79 is divisible by 71 + 8*1 = 79 |
|
|
A number is divisible by 99 if and only if the sum of numbers that form groups of two digits (starting with ones) is divisible by 99. |
For example, 12573 is divisible by 99, since 1 + 25 + 73 = 99 is divisible by 99. |
|
|
at 125 |
A number consisting of the last three digits is divisible by 125 |
……………375 |
