Signs of division by 1. Start in science

• that an integer a is divisible by an integer b is necessary and sufficient that a is divisible by the modulus of b;
• if in the equality a=s+t all terms, except for some one, are divisible by some integer b, then this one term is also divisible by b.

Now you can do proof of divisibility by 9. For convenience, we rewrite this feature as a necessary and sufficient condition for divisibility by 9.

Theorem.

For an integer a to be divisible by 9, it is necessary and sufficient that the sum of the digits in the record of the number a is divisible by 9.

Proof.

For a=0 the theorem is obvious.

For a , non-zero, the modulus of a is a natural number, so it can be represented as a sum, which we showed before the theorem. The expression contains a factor 9 , and the sum in brackets is a natural number for any a n , a n−1 , …, a 1 , therefore, due to the divisibility properties, the specified expression is divisible by 9 .

We proceed to the proof of sufficiency. Let's prove that if the sum of the digits of the number a (which we denoted as A) is divisible by 9, then the number a is divisible by 9.

If A is divisible by 9 , then it follows from equality and the second divisibility property indicated before the theorem that the module a is divisible by 9 , whence, by virtue of the first divisibility property indicated before the theorem, it follows that a is also divisible by 9 . This proves the sufficiency.

Let's move on to the proof of necessity. We prove that if an integer a is divisible by 9 , then the sum of its digits is divisible by 9 .

If a is divisible by 9, then the modulus of a is also divisible by 9 (according to the first divisibility property specified before the theorem). Then it follows from the equality and the second specified property of divisibility that A is divisible by 9. This proves the necessity.

This completes the proof of the test for divisibility by 9.

Other cases of divisibility by 9

In this paragraph, we want to touch on examples of proving divisibility by 9 when the number is given as the value of a literal expression for some values ​​of the variable.

Example.

Is 10 n −1 divisible by 9 for any natural n ?

Solution.

It is quite obvious that . The sum of the digits of the number is 9 n, and 9 n is divisible by 9, therefore, by the criterion of divisibility by 9, we can talk about the divisibility of 10 n −1 by 9 for any natural n.

Answer:

Yes, it is shared.

However, in many cases, it is not possible to use the sign of divisibility by 9, as well as perform division by 9. In such cases, it is logical to try to represent the original expression as a product of several integer factors, one of which is divisible by 9 . We show two ways to obtain such a product.

Sometimes it allows you to get a product of the desired type. Consider an example.

Example.

Is it divisible by 9 for any natural n ?

Solution.

Let's represent 4 as 3+1 , apply Newton's binomial formula and make transformations:

At n=1 we have , and 9 is divisible by 9 . For natural n , greater than one, in the resulting sum, 9 can be taken out of brackets, and we will come to the product. This product is divisible by 9 because it contains the factor 9 , and the value of the expression in brackets for n>1 is a natural number. Thus, is divisible by 9 for any natural n.

Answer:

Yes, share

When the original expression with the variable n is given as a polynomial, then this approach can be tried. If we prove that for n=9 m , n=9 m+1 , …, n=9 m+8 , where m is an integer, the original expression is divisible by 9 , then this will prove the divisibility of the original expression by 9 for any integer n.

Example.

Prove that is divisible by 9 for any integer n.

Solution.

For convenience, we factorize the expression :

Let m is an integer.

At n=9 m we get . The resulting product is divisible by 9 since the factor (9·m) 2 is obviously divisible by 9 .

At n=9 m+1 we have

The resulting product is divisible by 9, since it contains a factor of 9.

Divisibility by 9 of the expression is shown similarly for n=9 m+2 , n=9 m+3 , …, n=9 m+8 .

This proves the divisibility by 9 of the value of the original expression for any integer n 9 .

Based on the assumption of the previous step, we will prove that is divisible by 9 for n=k+1 .

We have

In the resulting difference is divisible by 9 since we assumed in the previous step that it is divisible by 9 ; and is also divisible by 9, since it contains a factor of 9. Consequently, the whole difference is divisible by 9, and hence the value of the expression for n=k+1 is divisible by 9.

So by the method of mathematical induction it was proved that it is divisible by 9 for any natural n.

Bibliography.

• Vilenkin N.Ya. etc. Mathematics. Grade 6: textbook for educational institutions.
• Vinogradov I.M. Fundamentals of number theory.
• Mikhelovich Sh.Kh. 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.

This article reveals the meaning of the sign of divisibility by 6. Its wording will be introduced with examples of solutions. Below we give a proof of the test for divisibility by 6 using some expressions as an example.

Sign of divisibility by 6, examples

The wording of the sign of divisibility by 6 includes the sign of divisibility by 2 and by 3: if the number ends with the numbers 0, 2, 4, 6, 8, and the sum of the digits is divisible without a remainder by 3, then such a number is divisible by 6; in the absence of at least one condition, the given number cannot be divided by 6. In other words, a number will be divisible by 6 when it is divisible by 2 and 3.

Applying the test for divisibility by 6 works in 2 steps:

• checking divisibility by 2, that is, the number must end in 2 for explicit divisibility by 2, if there are no digits 0, 2, 4, 6, 8 at the end of the number, division by 6 is impossible;
• checking divisibility by 3, and the check is made by dividing the sum of the digits of the number by 3 without a remainder, which means that the whole number can be divisible by 3; based on the previous paragraph, it can be seen that the whole number is divisible by 6, since the conditions for dividing by 3 and by 2 are met.

Example 1

Check if the number 8813 is divisible by 6 ?

Solution

Obviously, to answer you need to pay attention to the last digit of the number. Since 3 is not divisible by 2, it follows that one condition is not met. We get that the given number is not divisible by 6.

Answer: No.

Example 2

Find out if it is possible to divide the number 934 by 6 without a remainder.

Solution

Answer: No.

Example 3

Check divisibility by 6 numbers − 7 269 708 .

Solution

Let's move on to the last digit. Since its value is 8 , the first condition is satisfied, that is, 8 is divisible by 2 . Let's pass to check on satisfiability of the second condition. To do this, add the digits of the given number 7 + 2 + 6 + 9 + 7 + 0 + 8 = 39. It can be seen that 39 is divisible by 3 without a remainder. That is, we get (39: 3 = 13) . Obviously, both conditions are met, which means that the given number is divisible by 6 without a remainder.

Answer: yes, share.

To check divisibility by 6 , you can directly divide by 6 without checking the divisibility criteria for it.

Proof of the test for divisibility by 6

Consider the proof of the test for divisibility by 6 with necessary and sufficient conditions.

Theorem 1

In order for an integer a to be divisible by 6 , it is necessary and sufficient that this number be divisible by 2 and 3 .

Proof 1

To begin with, it is necessary to prove that the divisibility of the number a by 6 determines its divisibility by 2 and by 3. Using the divisibility property: if an integer is divisible by b , then the product of m a with m being an integer is also divisible by b .

It follows that when dividing a by 6, one can use the divisibility property to represent the equality as a = 6 q, where q is some integer. Such a record of the product suggests that the presence of a factor gives a guarantee of division by 2 and by 3. The need has been proven.

To fully prove divisibility by 6, one must prove sufficiency. To do this, you need to prove that if a number is divisible by 2 and by 3, then it is also divisible by 6 without a remainder.

It is necessary to apply the fundamental theorem of arithmetic. If the product of several positive integer factors not equal to one is divisible by a prime number p , then at least one factor is divisible by p .

We have that the integer a is divisible by 2, then there is such a number qwhen a = 2 · q. The same expression is divisible by 3, where 2 · q is divisible by 3. Obviously 2 is not divisible by 3. It follows from the theorem that q must be divisible by 3. From here we get that there is an integer q 1 , where q = 3 · q 1 . Hence, the resulting inequality of the form a = 2 q = 2 3 q 1 = 6 q 1 says that the number a will be divisible by 6 . Sufficiency has been proven.

Other cases of divisibility by 6

In this section, we consider ways to prove divisibility by 6 with variables. Such cases provide a different method of solution. We have a statement: if one of the integer factors in the product is divisible by a given number, then the entire product is divisible by this number. In other words, when a given expression is presented as a product, at least one of the factors is divisible by 6, then the entire expression will be divisible by 6.

Such expressions are easier to solve by substituting Newton's binomial formula.

Example 4

Determine if the expression 7 n - 12 n + 11 is divisible by 6 .

Solution

Let's represent the number 7 as the sum of 6 + 1 . From here we get a record of the form 7 n - 12 n + 11 = (6 + 1) n - 12 n + 11 . We apply Newton's binomial formula. After transformations, we have

7 n - 12 n + 11 = (6 + 1) n - 12 n + 11 = = (C n 0 6 n + C n 1 6 n - 1 + . . . + + C n n - 2 6 2 1 n - 2 + C n n - 1 6 1 n - 1 + C n n 1 n) - 12 n + 11 = = (6 n + C n 1 6 n - 1 + . . . + C n n - 2 6 2 + n 6 + 1) - 12 n + 11 = = 6 n + C n 1 6 n - 1 + . . . + C n n - 2 6 2 - 6 n + 12 = = 6 (6 n - 1 + C n 1 6 n - 2 + . . . + C n n - 2 6 1 - n + 2)

The resulting product is divisible by 6, because one of the factors is equal to 6. It follows that n can be any whole natural number, and the given expression is divisible by 6 .

Answer: Yes.

When an expression is specified using a polynomial, then conversions must be made. We see that it is required to resort to the decomposition of the polynomial into factors. we get that the variable n takes the form and is written as n = 6 m , n = 6 m + 1 , n = 6 m + 2 , … , n = 6 m + 5 , the number m is an integer. If divisibility for every n makes sense, then the divisibility of a given number by 6 for any value of the integer n will be proven.

Example 5

Prove that for any value of the integer n, the expression n 3 + 5 n is divisible by 6 .

Solution

First, let's factorize the given expression and get that n 3 + 5 n = n · (n 2 + 5) . If n \u003d 6 m, then n (n 2 + 5) \u003d 6 m (36 m 2 + 5) . Obviously, the presence of a factor of 6 means that the expression is divisible by 6 for any integer value m .

If n = 6 m + 1, we get

n (n 2 + 5) = (6 m + 1) 6 m + 1 2 + 5 = = (6 m + 1) (36 m 2 + 12 m + 1 + 5) = = (6 m + 1) 6 (6 m 2 + 2 m + 1)

The product will be divisible by 6 because it has a factor of 6 .

If n = 6 m + 2, then

n (n 2 + 5) = (6 m + 2) 6 m + 2 2 + 5 = = 2 (3 m + 1) (36 m 2 + 24 m + 4 + 5) = = 2 (3 m + 1) 3 (12 m 2 + 8 m + 3) = = 6 (3 m + 1) (12 m 2 + 8 m + 3)

The expression will be divisible by 6 because there is a factor of 6 in the notation.

The same holds true for n = 6 · m + 3 , n = 6 · m + 4 and n = 6 · m + 5 . When substituting, we will come to the conclusion that for any integer value of m, these expressions will be divisible by 6 . It follows that the given expression will be divisible by 6 for any integer value of n.

Now consider the example of a solution using the method of mathematical induction. A decision will be made according to the condition of the first example.

Example 6

Prove that an expression of the form 7 n - 12 n + 11 will be divisible by 6 , where it will take any integer values ​​of the expression.

Solution

We will solve this example by the method of mathematical induction. The algorithm is executed strictly step by step.

Let's check the divisibility of the expression by 6 for n = 1. Then we get an expression like 7 1 - 12 · 1 + 11 = 6 . Obviously 6 is divisible by itself.

Let's take n = k in the original expression. When it is divisible by 6 , then we can assume that 7k - 12k + 11 will be divisible by 6 .

Let's move on to the proof of division by 6 expressions of the form 7 n - 12 n + 11 for n = k + 1 . From here we get that it is necessary to prove the divisibility of the expression 7 k + 1 - 12 (k + 1) + 11 by 6, and it should be taken into account that 7 k - 12 k + 11 is divisible by 6 . Let's transform the expression and learn that

7 k + 1 - 12 (k + 1) + 11 = 7 7 k - 12 k - 1 = = 7 (7 k - 12 k + 11) + 72 k - 78 = = 7 (7 k - 12k + 11) + 6 (12k - 13)

Obviously, the first term will be divisible by 6, because 7k - 12k + 11 is divisible by 6. The second term is also divisible by 6, because one of the factors is 6. From this we conclude that all conditions are met, which means that the entire amount will be divisible by 6.

The method of mathematical induction proves that a given expression of the form 7 n - 12 n + 11 will be divisible by 6 when n takes the value of any natural number.

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A series of articles on the signs of divisibility continues sign of divisibility by 3. This article first gives the formulation of the criterion for divisibility by 3, and gives examples of the application of this criterion in finding out which of the given integers are divisible by 3 and which are not. Further, the proof of the divisibility test by 3 is given. Approaches to establishing the divisibility by 3 of numbers given as the value of some expression are also considered.

Page navigation.

Sign of divisibility by 3, examples

Let's start with formulations of the test for divisibility by 3: an integer is divisible by 3 if the sum of its digits is divisible by 3 , if the sum of its digits is not divisible by 3 , then the number itself is not divisible by 3 .

From the above formulation it is clear that the sign of divisibility by 3 cannot be used without the ability to perform. Also, for the successful application of the sign of divisibility by 3, you need to know that of all the numbers 3, 6 and 9 are divisible by 3, and the numbers 1, 2, 4, 5, 7 and 8 are not divisible by 3.

Now we can consider the simplest examples of applying the test for divisibility by 3. Find out if the number −42 is divisible by 3. To do this, we calculate the sum of the digits of the number −42, it is equal to 4+2=6. Since 6 is divisible by 3, then, by virtue of the divisibility criterion by 3, it can be argued that the number −42 is also divisible by 3. But the positive integer 71 is not divisible by 3, since the sum of its digits is 7+1=8, and 8 is not divisible by 3.

Is 0 divisible by 3? To answer this question, the test for divisibility by 3 is not needed, here we need to recall the corresponding divisibility property, which states that zero is divisible by any integer. So 0 is divisible by 3 .

In some cases, to show that a given number has or does not have the ability to be divisible by 3, the test for divisibility by 3 has to be applied several times in a row. Let's take an example.

Example.

Show that the number 907444812 is divisible by 3.

Solution.

The sum of the digits of 907444812 is 9+0+7+4+4+4+8+1+2=39 . To find out if 39 is divisible by 3 , we calculate its sum of digits: 3+9=12 . And to find out if 12 is divisible by 3, we find the sum of the digits of the number 12, we have 1+2=3. Since we got the number 3, which is divisible by 3, then, due to the sign of divisibility by 3, the number 12 is divisible by 3. Therefore, 39 is divisible by 3, since the sum of its digits is 12, and 12 is divisible by 3. Finally, 907333812 is divisible by 3 because the sum of its digits is 39 and 39 is divisible by 3.

To consolidate the material, we will analyze the solution of another example.

Example.

Is the number −543205 divisible by 3?

Solution.

Let's calculate the sum of the digits of this number: 5+4+3+2+0+5=19 . In turn, the sum of the digits of the number 19 is 1+9=10 , and the sum of the digits of the number 10 is 1+0=1 . Since we got the number 1, which is not divisible by 3, it follows from the criterion of divisibility by 3 that 10 is not divisible by 3. Therefore, 19 is not divisible by 3, because the sum of its digits is 10, and 10 is not divisible by 3. Therefore, the original number −543205 is not divisible by 3, since the sum of its digits, equal to 19, is not divisible by 3.

Answer:

No.

It is worth noting that the direct division of a given number by 3 also allows us to conclude whether the given number is divisible by 3 or not. By this we want to say that division should not be neglected in favor of the sign of divisibility by 3. In the last example, 543205 times 3 , we would make sure that 543205 is not even divisible by 3 , from which we could say that −543205 is not divisible by 3 either.

Proof of the test for divisibility by 3

The following representation of the number a will help us prove the sign of divisibility by 3. Any natural number a we can , after which it allows us to get a representation of the form , where a n , a n−1 , …, a 0 are the digits from left to right in the notation of the number a . For clarity, we give an example of such a representation: 528=500+20+8=5 100+2 10+8 .

Now let's write a number of fairly obvious equalities: 10=9+1=3 3+1 , 100=99+1=33 3+1 , 1 000=999+1=333 3+1 and so on.

Substituting into equality a=a n 10 n +a n−1 10 n−1 +…+a 2 10 2 +a 1 10+a 0 instead of 10 , 100 , 1 000 and so on expressions 3 3+1 , 33 3+1 , 999+1=333 3+1 and so on, we get
.

And allow the resulting equality to be rewritten as follows:

Expression is the sum of the digits of a. Let's denote it for brevity and convenience by the letter A, that is, let's take . Then we get a representation of the number a of the form , which we will use in proving the test for divisibility by 3 .

Also, to prove the test for divisibility by 3, we need the following properties of divisibility:

• that an integer a is divisible by an integer b is necessary and sufficient that a is divisible by the modulus of b;
• if in the equality a=s+t all terms, except for some one, are divisible by some integer b, then this one term is also divisible by b.

Now we are fully prepared and can carry out proof of divisibility by 3, for convenience, we formulate this feature as a necessary and sufficient condition for divisibility by 3 .

Theorem.

For an integer a to be divisible by 3, it is necessary and sufficient that the sum of its digits is divisible by 3.

Proof.

For a=0 the theorem is obvious.

If a is different from zero, then the modulus of a is a natural number, then the representation is possible, where is the sum of the digits of the number a.

Since the sum and product of integers is an integer, then is an integer, then by definition of divisibility, the product is divisible by 3 for any a 0 , a 1 , …, a n .

If the sum of the digits of the number a is divisible by 3, that is, A is divisible by 3, then, due to the divisibility property indicated before the theorem, it is divisible by 3, therefore, a is divisible by 3. This proves the sufficiency.

If a is divisible by 3, then it is divisible by 3, then due to the same divisibility property, the number A is divisible by 3, that is, the sum of the digits of the number a is divisible by 3. This proves the necessity.

Other cases of divisibility by 3

Sometimes integers are not specified explicitly, but as the value of some given value of the variable. For example, the value of an expression for some natural n is a natural number. It is clear that with this assignment of numbers, direct division by 3 will not help to establish their divisibility by 3, and the sign of divisibility by 3 will not always be able to be applied. Now we will consider several approaches to solving such problems.

The essence of these approaches is to represent the original expression as a product of several factors, and if at least one of the factors is divisible by 3, then, due to the corresponding property of divisibility, it will be possible to conclude that the entire product is divisible by 3.

Sometimes this approach allows you to implement. Let's consider an example solution.

Example.

Is the value of the expression divisible by 3 for any natural n ?

Solution.

The equality is obvious. Let's use Newton's binomial formula:

In the last expression, we can take 3 out of brackets, and we get . The resulting product is divisible by 3, since it contains a factor 3, and the value of the expression in brackets for natural n is a natural number. Therefore, is divisible by 3 for any natural n.

Answer:

Yes.

In many cases, proving divisibility by 3 allows . Let's analyze its application in solving an example.

Example.

Prove that for any natural n the value of the expression is divisible by 3 .

Solution.

For the proof, we use the method of mathematical induction.

At n=1 the value of the expression is , and 6 is divisible by 3 .

Suppose the value of the expression is divisible by 3 when n=k , that is, divisible by 3 .

Taking into account that it is divisible by 3 , we will show that the value of the expression for n=k+1 is divisible by 3 , that is, we will show that is divisible by 3.

Let's make some transformations:

The expression is divided by 3 and the expression is divisible by 3, so their sum is divisible by 3.

So the method of mathematical induction proved divisibility by 3 for any natural n.

Let's show one more approach to the proof of divisibility by 3 . If we show that for n=3 m , n=3 m+1 and n=3 m+2 , where m is an arbitrary integer, the value of some expression (with variable n ) is divisible by 3 , then this will prove divisibility of the expression by 3 for any integer n . Consider this approach when solving the previous example.

Thus, for any natural n is divisible by 3.

Answer:

Yes.

Bibliography.

• Vilenkin N.Ya. etc. Mathematics. Grade 6: textbook for educational institutions.
• Vinogradov I.M. Fundamentals of number theory.
• Mikhelovich Sh.Kh. Number theory.
• Kulikov L.Ya. and others. Collection of problems in algebra and number theory: Textbook for students of fiz.-mat. specialties of pedagogical institutes.

Let's start considering the topic "The sign of divisibility by 3". Let's start with the formulation of the sign, we will give the proof of the theorem. Then we will consider the main approaches to establishing the divisibility by 3 numbers, the value of which is given by some expression. The section provides an analysis of the solution of the main types of problems based on the use of the criterion of divisibility by 3 .

Sign of divisibility by 3, examples

The sign 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 an integer 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 by adding natural numbers.

Now let's look at examples of applying the divisibility criterion by 3.

Example 1

Is 42 divisible by 3?

Solution

In order to answer this question, let's add up all the numbers that make up the number - 42: 4 + 2 = 6.

Answer: according to the divisibility criterion, since the sum of the digits included in the rise of 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 the solution of which it is necessary to resort to the criterion of divisibility by 3 several times.

Example 2

Show that the number 907 444 812 is divisible by 3.

Solution

Let's find the sum of all the digits that form the record of the original number: 9 + 0 + 7 + 4 + 4 + 4 + 8 + 1 + 2 = 39 . Now we need to determine if the number 39 is divisible by 3. Once again, add the numbers that make up this number: 3 + 9 = 12 . It remains for us to carry out the addition of the numbers again in order 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 it 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, let's find the result of one more addition: 1 + 0 = 1 .
Answer: 1 is not divisible by 3, so the original number is not divisible by 3 either.

In order to determine whether a given number is divisible by 3 without a remainder, we can divide the given number by 3. If we divide the number − 543 205 from the above example with a column of three, then in the answer we will not get an integer. This also means exactly that − 543 205 is not divisible by 3.

Proof of the test for divisibility by 3

Here we need the following skills: decomposing a number into digits and the rule for multiplying 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 the number.

Here is an example using a specific number: 528 = 500 + 20 + 8 = 5 100 + 2 10 + 8.

Let's write a series of equalities: 10 = 9 + 1 = 3 3 + 1, 100 = 99 + 1 = 33 3 + 1, 1000 = 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.

So we came to 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

And now we 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 + . . . + a2 + a1 + a0

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 number representation is a = 3 33 . . . 3 a n + . . . + 33 · a 2 + 3 · a 1 + A takes a form that will be convenient for us to prove the test for divisibility by 3 .

Definition 1

Now remember the following properties of divisibility:

• a 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 divisible by the modulus of the number b ;
• if in equality a = s + t all terms, except for some one, are divisible by some integer b, then this one term is also divisible by b.

We have laid the foundation for proving the test for divisibility by 3. Now let us formulate this criterion in the form of a theorem and prove it.

Theorem 1

In order to assert that an integer a is divisible by 3, it is necessary and sufficient for us that the sum of the digits that form the record of the number a is divisible by 3.

Proof 1

If we take the value a = 0, then the theorem is obvious.

If we take a number a other than zero, then the absolute value of 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, by virtue of the divisibility property indicated before the theorem, a is divisible by 3 , hence, a divided by 3 . This proves the sufficiency.

If a divided by 3 , then a is 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 the number a divided by 3 . This proves the necessity.

Other cases of divisibility by 3

Integers can be given as the value of some expression that contains a variable, given a certain value of that variable. So, for some natural 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 of whether a number is divisible by 3 . Applying the divisibility test to 3 can also be difficult. Consider examples of such problems and analyze the methods for solving them.

Several approaches can be applied to solve such problems. The essence of one of them is as follows:

• represent the original expression as a product of several factors;
• find out if at least one of the factors can be divisible by 3 ;
• based on the divisibility property, we conclude that the entire product is divisible by 3 .

In the course of the solution, 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 for any natural n?

Solution

Let's write the equality 4 n + 3 n - 4 = (3 + 1) n + 3 n - 4 . We apply the Newton binomial formula of the Newton binomial:

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 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 a multiplier 3 , and the value of the expression in brackets for natural n is a natural number. This allows us to assert that the resulting product and the original expression 4 n + 3 n - 1 is divisible by 3 .

Answer: Yes.

We can also apply the method of mathematical induction.

Example 5

Prove using the method of mathematical induction that for any natural
n the value of the expression n n 2 + 5 is divisible by 3 .

Solution

Find the value of the expression n n 2 + 5 for n=1: 1 1 2 + 5 = 6 . 6 is divisible by 3 .

Now suppose that the value of the expression n n 2 + 5 for 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 .

Given that k k 2 + 5 is divisible by 3 , let us show that the value of the expression n n 2 + 5 for n=k+1 divided by 3 , that is, we will show that k + 1 k + 1 2 + 5 is divisible by 3 .

Let's do 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 divisible by 3 and the expression 3 k 2 + k + 2 is divisible by 3 , so their sum is divisible by 3 .

So we proved that the value of the expression n (n 2 + 5) is divisible by 3 for any natural n .

Let us now analyze the approach to the proof of divisibility by 3 , which is based on the following algorithm of actions:

• we show that the value of this expression with the variable n for n = 3 m , n = 3 m + 1 and n = 3 m + 2, Where m is 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 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 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 got contains the multiplier 3 , so the product itself is divisible by 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 n .

Example 7

Is it divided into 3 the value of the expression 10 3 n + 10 2 n + 1 for some natural 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 is divisible by 3 for any natural n.

Answer: Yes

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Let's proceed to the topic "Divisibility by 4". Here we give the formulation of the sign, carry out its proof, and consider the main examples of problems. At the end of the section, we collected information about approaches that can be used in cases where we need to prove the divisibility of numbers by 4 given by a literal expression.

Sign of divisibility by 4, examples

We can go the easy way and divide a single-digit natural number by 4 in order to check if this number is divisible by 4 without a remainder. You can do the same with two-digit, three-digit, etc. numbers. However, the larger the numbers become, the more difficult it is to perform operations with them in order to check their divisibility by 4.

It becomes much easier to use the divisibility test by 4 . It involves checking the divisibility of one or two last digits of an integer by 4. What does it mean? This means that some number a is divisible by 4 if one or two of the rightmost digits in the record of the number a are divisible by 4 . If the number composed of the two rightmost digits in the record of the number a is not divisible by 4 without a remainder, then the number a is not divisible by 4 without a remainder.

Example 1

Which of the numbers 98028, 7612 and 999 888 777 divisible by 4?

Solution

Rightmost digits of numbers − 98028, 7612 make up the numbers 28 and 12, which are divisible by 4 without a remainder. This means that the integers − 98028, 7612 are divisible by 4 without a remainder.

The last two digits of a number 999 888 777 form the number 77, which is not divisible by 4 without a remainder. This means that the original number is not divisible by 4 without a remainder.

Answer:− 98 028 and 7 612 .

If the penultimate digit in the number entry is 0, then we need to discard this zero and look at the remaining rightmost digit in the entry. It turns out that we replace the two digits 01 with 1 . And already by one remaining digit, we conclude whether the original number is divisible by 4.

Example 2

Is the number divisible 75 003 And − 88 108 for 4?

Solution

The last two digits of the number 75 003 - see 03 . If we discard zero, then we are left with the number 3, which is not divisible by 4 without a remainder. This means that the original number 75 003 is not divisible by 4.

Now take the last two digits of the number − 88 108 . This is 08 , of which we should leave only the last digit 8 . 8 is divisible by 4 without a remainder.

This means that the original number − 88 108 we can divide by 4 without a remainder.

Answer: 75 003 is not divisible by 4, but − 88 108 - shares.

Numbers that have two zeros at the end of the record are also divisible by 4 without a remainder. For example, 100 divided by 4 gives 25. The rule of multiplying a number by 100 allows us to prove the veracity of this statement.

Let us represent an arbitrarily chosen multi-valued number a , whose entry on the right ends with two zeros, as a product a 1 100, where the number a 1 is obtained from the number a if two zeros are discarded in its notation on the right. For example, 486700 = 4867 100 .

Work a 1 100 contains a factor of 100 , which is divisible by 4 . This means that the whole given product is divisible by 4.

Proof of the test for divisibility by 4

Imagine any natural number a in the form of equality a = a1 100 + a0, in which the number a 1 is a number a, from the record of which the last two digits were removed, and the number a 0- these are the two rightmost digits from the number entry a. If you use specific natural numbers, then the equality will look like undefined. For one and two digit numbers a = a0.

Definition 1

Now let's turn to the properties of divisibility:

• modulus division a modulo b is necessary and sufficient for an integer a divided by an integer b ;
• if in the equation a = s + t all terms except one are divisible by some integer b, then this remaining term is also divisible by the number b.

Now, having refreshed in memory the necessary properties of divisibility, we reformulate the proof of the test for divisibility by 4 in the form of a necessary and sufficient condition for divisibility by 4 .

Theorem 1

The division of the last two digits in the notation of the number a by 4 is a necessary and sufficient condition for the divisibility of the integer a by 4.

Proof 1

If we assume that a = 0, then the theorem does not need proof. For all other integers a we will use the modulus of a , which is a positive number: a = a 1 100 + a 0

Considering that the work a 1 100 is always divisible by 4, and also taking into account the divisibility properties that we gave above, we can make the following statement: if the number a is divisible by 4, then the modulus of the number a is also divisible by 4, then from the equality a = a 1 100 + a 0 it follows that a 0 is divisible by 4. So we proved the necessity.

From the equality a = a 1 100 + a 0 it follows that the modulus a is divisible by 4 . This means that the number a itself is divisible by 4. Thus we have proved sufficiency.

Other cases of divisibility by 4

Consider the cases when we need to set the divisibility by 4 of an integer given by some expression, the value of which must be calculated. To do this, we can go the following way:

• present the original expression as a product of several factors, one of which will be divisible by 4;
• draw a conclusion based on the property of divisibility that the entire original expression is divisible by
  4 .

Newton's binomial formula often helps in solving the problem.

Example 3

Is the value of the expression 9 n - 12 n + 7 divisible by 4 for some natural n?

Solution

We can represent 9 as the sum of 8 + 1 . This gives us the opportunity to apply Newton's binomial formula:

9 n - 12 n + 7 = 8 + 1 n - 12 n + 7 = = C n 0 8 n + C n 1 8 n - 1 1 + . . . + C n n - 2 8 2 1 n - 2 + C n n - 1 8 1 n - 1 + C n n 1 n - - 12 n + 7 = = 8 n + C n 1 8 n - 1 · 1 + . . . + C n n - 2 8 2 + n 8 + 1 - - 12 n + 7 = = 8 n + C n 1 8 n - 1 1 + . . . + C n n - 2 8 2 - 4 n + 8 = = 4 2 8 n - 1 + 2 C n 1 8 n - 2 + . . . + 2 C n n - 2 8 1 - n + 2

The product that we got during the transformations contains a factor of 4, and the expression in brackets is a natural number. This means that this product can be divided by 4 without a remainder.

We can say that the original expression 9 n - 12 n + 7 is divisible by 4 for any natural n .

Answer: Yes.

We can also apply the method of mathematical induction to the solution of the problem. In order not to divert your attention to minor details of the analysis of the solution, let's take the previous example.

Example 4

Prove that 9 n - 12 n + 7 is divisible by 4 for any natural n .

Solution

Let's start by establishing that with the value n=1 value of expression 9 n - 12 n + 7
can be divided by 4 without a remainder.

We get: 9 1 - 12 1 + 7 = 4. 4 is divisible by 4 without a remainder.

Now we can assume that with the value n=k expression value
9 n - 12 n + 7 will be divisible by 4 . In fact, we will work with the expression 9 k - 12 k + 7 , which must be divisible by 4 .

We need to prove that 9 n - 12 n + 7 for n=k+1 will be divisible by 4, given that 9 k - 12 k + 7 is divisible by 4:

9 k + 1 - 12 (k + 1) + 7 = 9 9 k - 12 k - 5 = 9 9 k - 12 k + 7 + 96 k - 68 = = 9 9 k - 12 k + 7 + 4 24k - 17

We have obtained a sum in which the first term 9 9 k - 12 k + 7 is divisible by 4 due to our assumption that 9 k - 12 k + 7 is divisible by 4 , and the second term 4 24 k - 17 contains factor 4 , in connection with which it is also divisible by 4 . This means that the whole sum is divisible by 4.

Answer: we proved that 9 n - 12 n + 7 is divisible by 4 for any natural value of n by mathematical induction.

We can use another approach to prove that some expression is divisible by 4 . This approach assumes:

• proof of the fact that the value of the given expression with variable n is divisible by 4 for n = 4 m , n = 4 m + 1 , n = 4 m + 2 and n = 4 m + 3, Where m is an integer;
• conclusion about the proof of the divisibility of the given expression by 4 for any integer n .

Example 5

Prove that the value of the expression n n 2 + 1 n + 3 n 2 + 4 for any integer n is divisible by 4.

Solution

If we assume that n = 4 m, we get:

4 m 4 m 2 + 1 4 m + 3 4 m 2 + 4 = 4 m 16 m 2 + 1 4 m + 3 4 4 m 2 + 1

The resulting product contains a factor of 4 , all other factors are integers. This gives us reason to assume that the entire product is divisible by 4.

If we assume that n = 4 m + 1, we get:

4 m + 1 4 m + 1 2 + 1 4 m + 1 + 3 4 m + 1 2 + 4 = = (4 m 1) + 4 m + 1 2 + 1 4 m + 1 4 m + 1 2 + 4

And again in the work that we received in the course of transformations,
contains a factor of 4 .

This means that the expression is divisible by 4 .

If we assume that n = 4 m + 2, then:

4 m + 2 4 m + 2 2 + 1 4 m + 2 + 3 4 m + 2 2 + 4 = = 2 2 m + 1 16 m 2 + 16 m + 5 (4 m + 5 ) 8 (2 m 2 + 2 m + 1)

Here in the product we got a factor 8, which can be divided by 4 without a remainder. This means that the whole product is divisible by 4.

If we assume that n = 4 m + 3, we get:

4 m + 3 4 m + 3 2 + 1 4 m + 3 + 3 4 m + 3 2 + 4 = = 4 m + 3 2 8 m 2 + 12 m + 5 2 2 m + 3 16 m 2 + 24 m + 13 = = 4 4 m + 3 8 m 2 + 12 m + 5 16 m 2 + 24 m + 13

The product contains a factor of 4, which means that it is divisible by 4 without a remainder.

Answer: we have proved that the original expression is divisible by 4 for any n .

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