What factors does it decompose into? Factoring a number into prime factors online

What does it mean to decompose into prime factors? How to do this? What can you learn from factoring a number into prime factors? The answers to these questions are illustrated with specific examples.

Definitions:

A number that has exactly two different divisors is called prime.

A number that has more than two divisors is called composite.

Expand natural number to factor means to represent it as a product of natural numbers.

To factor a natural number into prime factors means to represent it as a product of prime numbers.

Notes:

In the expansion of a prime number, one of the factors equal to one, and the other - to this number itself.
It makes no sense to talk about factoring unity.
A composite number can be factored into factors, each of which is different from 1.

	Let's factor the number 150. For example, 150 is 15 times 10. 15 is a composite number. It can be factored into prime factors of 5 and 3. 10 is a composite number. It can be factored into prime factors of 5 and 2. By writing their decompositions into prime factors instead of 15 and 10, we obtained the decomposition of the number 150.
	The number 150 can be factorized in another way. For example, 150 is the product of the numbers 5 and 30. 5 is a prime number. 30 is a composite number. It can be thought of as the product of 10 and 3. 10 is a composite number. It can be factored into prime factors of 5 and 2. We obtained the factorization of 150 into prime factors in a different way.
	Note that the first and second expansions are the same. They differ only in the order of the factors. It is customary to write factors in ascending order.
Every composite number can be factorized into prime factors in a unique way, up to the order of the factors.

When factoring large numbers into prime factors, use column notation:

The smallest prime number that is divisible by 216 is 2.

Divide 216 by 2. We get 108.

The resulting number 108 is divided by 2.

Let's do the division. The result is 54.

According to the test of divisibility by 2, the number 54 is divisible by 2.

After dividing, we get 27.

The number 27 ends with the odd digit 7. It

Not divisible by 2. The next prime number is 3.

Divide 27 by 3. We get 9. Least prime

The number that 9 is divided by is 3. Three is itself prime number, it is divisible by itself and by one. Let's divide 3 by ourselves. In the end we got 1.

A number is divisible only by those prime numbers that are part of its decomposition.
A number is divisible only into those composite numbers whose decomposition into prime factors is completely contained in it.

Let's look at examples:

4900 is divisible by the prime numbers 2, 5 and 7 (they are included in the expansion of the number 4900), but is not divisible by, for example, 13.

11 550 75. This is so because the decomposition of the number 75 is completely contained in the decomposition of the number 11550.

The result of division will be the product of factors 2, 7 and 11.

11550 is not divisible by 4 because there is an extra two in the expansion of four.

Find the quotient of dividing the number a by the number b, if these numbers are decomposed into prime factors as follows: a=2∙2∙2∙3∙3∙3∙5∙5∙19; b=2∙2∙3∙3∙5∙19

The decomposition of the number b is completely contained in the decomposition of the number a.

The result of dividing a by b is the product of the three numbers remaining in the expansion of a.

So the answer is: 30.

References

Vilenkin N.Ya., Zhokhov V.I., Chesnokov A.S., Shvartsburd S.I. Mathematics 6. - M.: Mnemosyne, 2012.
Merzlyak A.G., Polonsky V.V., Yakir M.S. Mathematics 6th grade. - Gymnasium. 2006.
Depman I.Ya., Vilenkin N.Ya. Behind the pages of a mathematics textbook. - M.: Education, 1989.
Rurukin A.N., Tchaikovsky I.V. Assignments for the mathematics course, grades 5-6. - M.: ZSh MEPhI, 2011.
Rurukin A.N., Sochilov S.V., Tchaikovsky K.G. Mathematics 5-6. A manual for 6th grade students at the MEPhI correspondence school. - M.: ZSh MEPhI, 2011.
Shevrin L.N., Gein A.G., Koryakov I.O., Volkov M.V. Mathematics: Textbook-interlocutor for grades 5-6 high school. - M.: Education, Mathematics Teacher Library, 1989.

Internet portal Matematika-na.ru ().
Internet portal Math-portal.ru ().

Homework

Vilenkin N.Ya., Zhokhov V.I., Chesnokov A.S., Shvartsburd S.I. Mathematics 6. - M.: Mnemosyne, 2012. No. 127, No. 129, No. 141.
Other tasks: No. 133, No. 144.

Every natural number, except one, has two or more divisors. For example, the number 7 is divisible without a remainder only by 1 and 7, that is, it has two divisors. And the number 8 has divisors 1, 2, 4, 8, that is, as many as 4 divisors at once.

What is the difference between prime and composite numbers?

Numbers that have more than two divisors are called composite numbers. Numbers that have only two divisors: one and the number itself are called prime numbers.

The number 1 has only one division, namely the number itself. One is neither a prime nor a composite number.

For example, the number 7 is prime and the number 8 is composite.

First 10 prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. The number 2 is the only even prime number, all other prime numbers are odd.

The number 78 is composite, since in addition to 1 and itself, it is also divisible by 2. When divided by 2, we get 39. That is, 78 = 2*39. In such cases, they say that the number was factored into factors of 2 and 39.

Any composite number can be decomposed into two factors, each of which is greater than 1. This trick will not work with a prime number. Such things.

Factoring a number into prime factors

As noted above, any composite number can be factorized into two factors. Let's take, for example, the number 210. This number can be decomposed into two factors 21 and 10. But the numbers 21 and 10 are also composite, let's decompose them into two factors. We get 10 = 2*5, 21=3*7. And as a result, the number 210 was decomposed into 4 factors: 2,3,5,7. These numbers are already prime and cannot be expanded. That is, we factored the number 210 into prime factors.

When factoring composite numbers into prime factors, they are usually written in ascending order.

It should be remembered that any composite number can be decomposed into prime factors and in a unique way, up to permutation.

Usually, when decomposing a number into prime factors, divisibility criteria are used.

Let's factor the number 378 into prime factors

We will write down the numbers, separating them with a vertical line. The number 378 is divisible by 2, since it ends in 8. When divided, we get the number 189. The sum of the digits of the number 189 is divisible by 3, which means the number 189 itself is divisible by 3. The result is 63.

The number 63 is also divisible by 3, according to divisibility. We get 21, the number 21 can again be divided by 3, we get 7. Seven is divided only by itself, we get one. This completes the division. To the right after the line are the prime factors into which the number 378 is decomposed.

378|2
189|3
63|3
21|3

Any composite number can be factorized into prime factors. There can be several methods of decomposition. Either method produces the same result.

How to factor a number into prime factors most in a convenient way? Let's look at how best to do this, using specific examples.

Examples. 1) Factor the number 1400 into prime factors.

1400 is divisible by 2. 2 is a prime number; there is no need to factor it. We get 700. Divide it by 2. We get 350. We also divide 350 by 2. The resulting number 175 can be divided by 5. The result is 35 - we divide it by 5 again. Total is 7. It can only be divided by 7. We get 1, division over.

The same number can be factorized differently:

It is convenient to divide 1400 by 10. 10 is not a prime number, so it needs to be factored into prime factors: 10=2∙5. The result is 140. We divide it again by 10=2∙5. We get 14. If 14 is divided by 14, then it should also be decomposed into a product of prime factors: 14=2∙7.

Thus, we again came to the same decomposition as in the first case, but faster.

Conclusion: when decomposing a number, it is not necessary to divide it only into prime factors. We divide by what is more convenient, for example, by 10. You just need to remember to decompose the compound divisors into simple factors.

2) Factor the number 1620 into prime factors.

The most convenient way to divide the number 1620 is by 10. Since 10 is not a prime number, we represent it as a product of prime factors: 10=2∙5. We got 162. It is convenient to divide it by 2. The result is 81. The number 81 can be divided by 3, but by 9 it is more convenient. Since 9 is not a prime number, we expand it as 9=3∙3. We get 9. We also divide it by 9 and expand it into the product of prime factors.

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This article gives answers to the question of factoring a number on a sheet. Let's consider general idea about decomposition with examples. Let us analyze the canonical form of the expansion and its algorithm. All will be considered alternative ways using divisibility signs and multiplication tables.

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What does it mean to factor a number into prime factors?

Let's look at the concept of prime factors. It is known that every prime factor is a prime number. In a product of the form 2 · 7 · 7 · 23 we have that we have 4 prime factors in the form 2, 7, 7, 23.

Factorization involves its representation in the form of products of primes. If we need to decompose the number 30, then we get 2, 3, 5. The entry will take the form 30 = 2 · 3 · 5. It is possible that the multipliers may be repeated. A number like 144 has 144 = 2 2 2 2 3 3.

Not all numbers are prone to decay. Numbers that are greater than 1 and are integers can be factored. Prime numbers, when factored, are only divisible by 1 and themselves, so it is impossible to represent these numbers as a product.

When z refers to integers, it is represented as a product of a and b, where z is divided by a and b. Composite numbers are factored using the fundamental theorem of arithmetic. If the number is greater than 1, then its factorization p 1, p 2, ..., p n takes the form a = p 1 , p 2 , … , p n . The decomposition is assumed to be in a single variant.

Canonical factorization of a number into prime factors

During expansion, factors can be repeated. They are written compactly using degrees. If, when decomposing the number a, we have a factor p 1, which occurs s 1 times and so on p n – s n times. Thus the expansion will take the form a=p 1 s 1 · a = p 1 s 1 · p 2 s 2 · … · p n s n. This entry is called the canonical factorization of a number into prime factors.

When expanding the number 609840, we get that 609 840 = 2 2 2 2 3 3 5 7 11 11, its canonical form will be 609 840 = 2 4 3 2 5 7 11 2. Using canonical expansion, you can find all the divisors of a number and their number.

To correctly factorize, you need to have an understanding of prime and composite numbers. The point is to obtain a sequential number of divisors of the form p 1, p 2, ..., p n numbers a , a 1 , a 2 , … , a n - 1, this makes it possible to get a = p 1 a 1, where a 1 = a: p 1 , a = p 1 · a 1 = p 1 · p 2 · a 2 , where a 2 = a 1: p 2 , … , a = p 1 · p 2 · … · p n · a n , where a n = a n - 1: p n. Upon receipt a n = 1, then the equality a = p 1 · p 2 · … · p n we obtain the required decomposition of the number a into prime factors. Note that p 1 ≤ p 2 ≤ p 3 ≤ … ≤ p n.

To find the smallest common divisors you need to use a table of prime numbers. This is done using the example of finding the smallest prime divisor of the number z. When taking prime numbers 2, 3, 5, 11 and so on, and dividing the number z by them. Since z is not a prime number, it should be taken into account that the smallest prime divisor will not be greater than z. It can be seen that there are no divisors of z, then it is clear that z is a prime number.

Example 1

Let's look at the example of the number 87. When it is divided by 2, we have that 87: 2 = 43 with a remainder of 1. It follows that 2 cannot be a divisor; division must be done entirely. When divided by 3, we get that 87: 3 = 29. Hence the conclusion is that 3 is the smallest prime divisor of the number 87.

When factoring into prime factors, you must use a table of prime numbers, where a. When factoring 95, you should use about 10 primes, and when factoring 846653, about 1000.

Let's consider the decomposition algorithm into prime factors:

finding the smallest factor of divisor p 1 of a number a by the formula a 1 = a: p 1, when a 1 = 1, then a is a prime number and is included in the factorization, when not equal to 1, then a = p 1 · a 1 and follow to the point below;
finding the prime divisor p 2 of a number a 1 by sequentially enumerating prime numbers using a 2 = a 1: p 2 , when a 2 = 1 , then the expansion will take the form a = p 1 p 2 , when a 2 = 1, then a = p 1 p 2 a 2 , and we move on to the next step;
searching through prime numbers and finding a prime divisor p 3 numbers a 2 according to the formula a 3 = a 2: p 3 when a 3 = 1 , then we get that a = p 1 p 2 p 3 , when not equal to 1, then a = p 1 p 2 p 3 a 3 and move on to the next step;
the prime divisor is found p n numbers a n - 1 by enumerating prime numbers with pn - 1, and also a n = a n - 1: p n, where a n = 1, the step is final, as a result we get that a = p 1 · p 2 · … · p n .

The result of the algorithm is written in the form of a table with decomposed factors with a vertical bar sequentially in a column. Consider the figure below.

The resulting algorithm can be applied by decomposing numbers into prime factors.

When factoring into prime factors, the basic algorithm should be followed.

Example 2

Factor the number 78 into prime factors.

Solution

In order to find the smallest prime divisor, you need to go through all the prime numbers in 78. That is 78: 2 = 39. Division without a remainder means this is the first simple divisor, which we denote as p 1. We get that a 1 = a: p 1 = 78: 2 = 39. We arrived at an equality of the form a = p 1 · a 1 , where 78 = 2 39. Then a 1 = 39, that is, we should move on to the next step.

Let's focus on finding the prime divisor p2 numbers a 1 = 39. You should go through the prime numbers, that is, 39: 2 = 19 (remaining 1). Since division with a remainder, 2 is not a divisor. When choosing the number 3, we get that 39: 3 = 13. This means that p 2 = 3 is the smallest prime divisor of 39 by a 2 = a 1: p 2 = 39: 3 = 13. We obtain an equality of the form a = p 1 p 2 a 2 in the form 78 = 2 3 13. We have that a 2 = 13 is not equal to 1, then we should move on.

The smallest prime divisor of the number a 2 = 13 is found by searching through numbers, starting with 3. We get that 13: 3 = 4 (remaining 1). From this we can see that 13 is not divisible by 5, 7, 11, because 13: 5 = 2 (rest. 3), 13: 7 = 1 (rest. 6) and 13: 11 = 1 (rest. 2). It can be seen that 13 is a prime number. According to the formula it looks like this: a 3 = a 2: p 3 = 13: 13 = 1. We found that a 3 = 1, which means the completion of the algorithm. Now the factors are written as 78 = 2 · 3 · 13 (a = p 1 · p 2 · p 3) .

Answer: 78 = 2 3 13.

Example 3

Factor the number 83,006 into prime factors.

Solution

The first step involves factoring p 1 = 2 And a 1 = a: p 1 = 83,006: 2 = 41,503, where 83,006 = 2 · 41,503.

The second step assumes that 2, 3 and 5 are not prime divisors for the number a 1 = 41,503, but 7 is a prime divisor, because 41,503: 7 = 5,929. We get that p 2 = 7, a 2 = a 1: p 2 = 41,503: 7 = 5,929. Obviously, 83,006 = 2 7 5 929.

Finding the smallest prime divisor of p 4 to the number a 3 = 847 is 7. It can be seen that a 4 = a 3: p 4 = 847: 7 = 121, so 83 006 = 2 7 7 7 121.

To find the prime divisor of the number a 4 = 121, we use the number 11, that is, p 5 = 11. Then we get an expression of the form a 5 = a 4: p 5 = 121: 11 = 11, and 83,006 = 2 7 7 7 11 11.

For number a 5 = 11 number p 6 = 11 is the smallest prime divisor. Hence a 6 = a 5: p 6 = 11: 11 = 1. Then a 6 = 1. This indicates the completion of the algorithm. The factors will be written as 83 006 = 2 · 7 · 7 · 7 · 11 · 11.

The canonical notation of the answer will take the form 83 006 = 2 · 7 3 · 11 2.

Answer: 83 006 = 2 7 7 7 11 11 = 2 7 3 11 2.

Example 4

Factor the number 897,924,289.

Solution

To find the first prime factor, search through the prime numbers, starting with 2. The end of the search occurs at the number 937. Then p 1 = 937, a 1 = a: p 1 = 897 924 289: 937 = 958 297 and 897 924 289 = 937 958 297.

The second step of the algorithm is to iterate over smaller prime numbers. That is, we start with the number 937. The number 967 can be considered prime because it is a prime divisor of the number a 1 = 958,297. From here we get that p 2 = 967, then a 2 = a 1: p 1 = 958 297: 967 = 991 and 897 924 289 = 937 967 991.

The third step says that 991 is a prime number, since it does not have a single prime factor that does not exceed 991. The approximate value of the radical expression is 991< 40 2 . Иначе запишем как 991 < 40 2 . This shows that p 3 = 991 and a 3 = a 2: p 3 = 991: 991 = 1. We find that the decomposition of the number 897 924 289 into prime factors is obtained as 897 924 289 = 937 967 991.

Answer: 897 924 289 = 937 967 991.

Using divisibility tests for prime factorization

To factor a number into prime factors, you need to follow an algorithm. When there are small numbers, it is permissible to use the multiplication table and divisibility signs. Let's look at this with examples.

Example 5

If it is necessary to factorize 10, then the table shows: 2 · 5 = 10. The resulting numbers 2 and 5 are prime numbers, so they are prime factors for the number 10.

Example 6

If it is necessary to decompose the number 48, then the table shows: 48 = 6 8. But 6 and 8 are not prime factors, since they can also be expanded as 6 = 2 3 and 8 = 2 4. Then the complete expansion from here is obtained as 48 = 6 8 = 2 3 2 4. The canonical notation will take the form 48 = 2 4 · 3.

Example 7

When decomposing the number 3400, you can use the signs of divisibility. In this case, the signs of divisibility by 10 and 100 are relevant. From here we get that 3,400 = 34 · 100, where 100 can be divided by 10, that is, written as 100 = 10 · 10, which means that 3,400 = 34 · 10 · 10. Based on the divisibility test, we find that 3 400 = 34 10 10 = 2 17 2 5 2 5. All factors are prime. The canonical expansion takes the form 3 400 = 2 3 5 2 17.

When we find prime factors, we need to use divisibility tests and multiplication tables. If you imagine the number 75 as a product of factors, then you need to take into account the rule of divisibility by 5. We get that 75 = 5 15, and 15 = 3 5. That is, the desired expansion is an example of the form of the product 75 = 5 · 3 · 5.

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