What prime number is divisible by 3. Division

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), but the sum of the digits is 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 whether a given number is divisible by a composite number, you need to factor that composite number into coprime factors, the signs of divisibility of which are known. Coprime numbers are numbers that have no common factors 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.

There are signs by which it is sometimes easy to find out, without actually dividing, whether a given number is divisible or not divisible by some other numbers.

Numbers that are divisible by 2 are called even. The number zero also refers to even numbers. All other numbers are called odd:

0, 2, 4, 6, 8, 10, 12, ... - even,
1, 3, 5, 7, 9, 11, 13, ... - odd.

Signs of divisibility

Test for divisibility by 2. A number is divisible by 2 if its last digit is even. For example, the number 4376 is divisible by 2, since the last digit (6) is even.

Test for divisibility by 3. Only those numbers whose sum of digits is divisible by 3 are divisible by 3. For example, the number 10815 is divisible by 3, since the sum of its digits 1 + 0 + 8 + 1 + 5 = 15 is divisible by 3.

Tests for divisibility by 4. A number is divisible by 4 if its last two digits are zeros or form a number that is divisible by 4. For example, the number 244500 is divisible by 4 because it ends with two zeros. The numbers 14708 and 7524 are divisible by 4 because the last two digits of these numbers (08 and 24) are divisible by 4.

Tests for divisibility by 5. Those numbers that end in 0 or 5 are divisible by 5. For example, the number 320 is divisible by 5, since the last digit is 0.

Test for divisibility by 6. A number is divisible by 6 if it is divisible by both 2 and 3. For example, the number 912 is divisible by 6 because it is divisible by both 2 and 3.

Tests for divisibility by 8. Divided by 8 are those numbers whose last three digits are zeros or form a number that is divisible by 8. For example, the number 27000 is divisible by 8, since it ends with three zeros. The number 63128 is divisible by 8 because the last three digits form the number (128), which is divisible by 8.

Divisibility test by 9. Only those numbers whose sum of digits is divisible by 9 are divisible by 9. For example, the number 2637 is divisible by 9, since the sum of its digits 2 + 6 + 3 + 7 = 18 is divisible by 9.

Signs of divisibility by 10, 100, 1000, etc. Those numbers that end in one zero, two zeros, three zeros, and so on are divided by 10, 100, 1000, and so on. For example, the number 3800 is divisible by 10 and 100.

The series of articles on divisibility criteria continues test of divisibility by 3. This article first gives a formulation of the test for divisibility by 3, and gives examples of using this test to find out which of the given integers are divisible by 3 and which are not. Below is a proof of the test for divisibility by 3. Also considered are approaches to establishing divisibility by 3 of numbers given as the value of some expression.

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Test for divisibility by 3, examples

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

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

Now we can consider the simplest examples of using the test of divisibility by 3. Let's find out whether 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, due to the divisibility test by 3, we can say 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, you won’t need the property of divisibility by 3; here you need to remember the corresponding property of divisibility, 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 of divisibility by 3 must be used several times in a row. Let's give an example.

Example.

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

Solution.

The sum of the digits of the number 907 444 812 is 9+0+7+4+4+4+8+1+2=39. To find out whether 39 is divisible by 3, let's calculate its sum of digits: 3+9=12. And to find out whether 12 is divisible by 3, we find the sum of the digits of the number 12, we have 1+2=3. Since we received the number 3, which is divisible by 3, then, by virtue of the divisibility test 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, 907,333,812 is divisible by 3, since the sum of its digits is 39, and 39 is divisible by 3.

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

Example.

Is −543,205 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 received the number 1, which is not divisible by 3, from the test of divisibility by 3 it follows that 10 is not divisible by 3. Therefore, 19 is not divisible by 3, since the sum of its digits is 10, and 10 is not divisible by 3. Therefore, the original number −543,205 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 directly dividing a given number by 3 also allows us to conclude whether a given number is divisible by 3 or not. By this we want to say that we should not neglect division in favor of the criterion of divisibility by 3. IN last example, 543,205 by 3, we would make sure that 543,205 is not divisible by 3, from which we could say that −543,205 is not divisible by 3.

Proof of the test of divisibility by 3

The following representation of the number a will help us prove the test of divisibility by 3. Any natural number a we can, after which it allows us to obtain 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 down 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, the expressions 3·3+1, 33·3+1, 999+1=333·3+1 and so on, we get
.

And they allow the resulting equality to be rewritten as follows:

Expression is the sum of the digits of the number a. For brevity and convenience, let us denote it by the letter A, that is, we accept . Then we get a representation of the number a of the form, which we will use to prove the test for divisibility by 3.

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

• For an integer a to be divisible by an integer b it is necessary and sufficient that a be divisible by the modulus of b;
• if in the equality a=s+t all terms except 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 criterion in the form of 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 the number 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 it is an integer, then, by the definition of divisibility, the product is divisible by 3 for any a 0, a 1, ..., a n.

If the sum of the digits of a 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. So the sufficiency is proven.

If a is divisible by 3, then it is divisible by 3, then, due to the same property of divisibility, the number A is divisible by 3, that is, the sum of the digits of the number a is divisible by 3. The necessity has been proven.

Other cases of divisibility by 3

Sometimes integers are not specified explicitly, but as the value of some given value variable. For example, the value of an expression for some natural number n is a natural number. It is clear that when specifying numbers in this way, direct division by 3 will not help to establish their divisibility by 3, and the test of divisibility by 3 cannot always 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 divisibility property, it will be possible to conclude that the entire product is divisible by 3.

Sometimes this approach allows you to implement it. Let's look at the example solution.

Example.

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

Solution.

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 divided by 3, since it contains a factor of 3, and the value of the expression in parentheses for natural n represents a natural number. Therefore, it is divisible by 3 for any natural number n.

Answer:

Yes.

In many cases, it is possible to prove divisibility by 3. Let's look at its application when solving an example.

Example.

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

Solution.

To prove this, we will use the method of mathematical induction.

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

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

Considering 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 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 by adding 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 need 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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