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Solved problems involving prime factors. Factoring a number

Factorize large number- not an easy task. Most people have trouble figuring out four or five digit numbers. To make the process easier, write the number above the two columns.

  • Let's factorize the number 6552.

  • Divide the given number by the smallest prime divisor (other than 1) that divides the given number without leaving a remainder. Write this divisor in the left column, and write the result of the division in the right column. As noted above, even numbers are easy to factor, since their smallest prime factor will always be the number 2 (odd numbers have smallest prime factors are different).

    • In our example, 6552 is an even number, so 2 is its smallest prime factor. 6552 ÷ 2 = 3276. Write 2 in the left column and 3276 in the right column.

  • Next, divide the number in the right column by the smallest prime factor (other than 1) that divides the number without leaving a remainder. Write this divisor in the left column, and in the right column write the result of the division (continue this process until there are no 1 left in the right column).

    • In our example: 3276 ÷ 2 = 1638. Write 2 in the left column, and 1638 in the right column. Next: 1638 ÷ 2 = 819. Write 2 in the left column, and 819 in the right column.

  • You got an odd number; For such numbers, finding the smallest prime divisor is more difficult. If you get an odd number, try dividing it by the smallest prime odd numbers: 3, 5, 7, 11.

    • In our example, you received an odd number 819. Divide it by 3: 819 ÷ 3 = 273. Write 3 in the left column and 273 in the right column.
    • When selecting divisors, try all prime numbers up to square root from the greatest divisor you found. If no divisor divides the number by an integer, then you most likely have a prime number and can stop calculating.

  • Continue the process of dividing numbers by prime factors until you are left with a 1 in the right column (if you get a prime number in the right column, divide it by itself to get a 1).

    • Let's continue the calculations in our example:
      • Divide by 3: 273 ÷ 3 = 91. There is no remainder. Write down 3 in the left column and 91 in the right column.
      • Divide by 3. 91 is divisible by 3 with a remainder, so divide by 5. 91 is divisible by 5 with a remainder, so divide by 7: 91 ÷ 7 = 13. No remainder. Write down 7 in the left column and 13 in the right column.
      • Divide by 7. 13 is divisible by 7 with a remainder, so divide by 11. 13 is divisible by 11 with a remainder, so divide by 13: 13 ÷ 13 = 1. There is no remainder. Write 13 in the left column and 1 in the right column. Your calculations are complete.

  • The left column shows the prime factors of the original number. In other words, when you multiply all the numbers in the left column, you will get the number written above the columns. If the same factor appears more than once in the list of factors, use exponents to indicate it. In our example, 2 appears 4 times in the list of multipliers; write these factors as 2 4 rather than 2*2*2*2.

    • In our example, 6552 = 2 3 × 3 2 × 7 × 13. You factored 6552 into prime factors (the order of the factors in this notation does not matter).

    • What's happened factorization? This is a way to turn an inconvenient and complex example into a simple and cute one.) A very powerful technique! It is found at every step in both elementary and higher mathematics.

    Such transformations in mathematical language are called identical transformations of expressions. For those who are not in the know, take a look at the link. There is very little, simple and useful.) The meaning of any identity transformation is a recording of the expression in another form while maintaining its essence.

    Meaning factorization extremely simple and clear. Right from the name itself. You may forget (or not know) what a multiplier is, but you can figure out that this word comes from the word “multiply”?) Factoring means: represent an expression in the form of multiplying something by something. May mathematics and the Russian language forgive me...) That's all.

    For example, you need to expand the number 12. You can safely write:

    So we presented the number 12 as a multiplication of 3 by 4. Please note that the numbers on the right (3 and 4) are completely different than on the left (1 and 2). But we understand perfectly well that 12 and 3 4 one and the same. The essence of the number 12 from transformation hasn't changed.

    Is it possible to decompose 12 differently? Easily!

    12=3·4=2·6=3·2·2=0.5·24=........

    The decomposition options are endless.

    Factoring numbers is a useful thing. It helps a lot, for example, when working with roots. But factoring algebraic expressions is not only useful, it is necessary! Just for example:

    Simplify:

    Those who do not know how to factor an expression rest on the sidelines. Those who know how - simplify and get:

    The effect is amazing, right?) By the way, the solution is quite simple. You'll see for yourself below. Or, for example, this task:

    Solve the equation:

    x 5 - x 4 = 0

    It is decided in the mind, by the way. Using factorization. We will solve this example below. Answer: x 1 = 0; x 2 = 1.

    Or, the same thing, but for the older ones):

    Solve the equation:

    In these examples I showed main purpose factorization: simplifying fractional expressions and solving some types of equations. Here's a rule of thumb to remember:

    If we have a scary fractional expression in front of us, we can try factoring the numerator and denominator. Very often the fraction is reduced and simplified.

    If we have an equation in front of us, where on the right there is zero, and on the left - I don’t understand what, we can try to factorize the left side. Sometimes it helps).

    Basic methods of factorization.

    Here they are, the most popular methods:

    4. Expansion of a quadratic trinomial.

    These methods must be remembered. Exactly in that order. Complex examples are checked for everything possible ways decomposition. And it’s better to check in order so as not to get confused... So let’s start in order.)

    1. Taking the common factor out of brackets.

    Simple and reliable way. Nothing bad comes from him! It happens either well or not at all.) That’s why he comes first. Let's figure it out.

    Everyone knows (I believe!) the rule:

    a(b+c) = ab+ac

    Or, more general view:

    a(b+c+d+.....) = ab+ac+ad+....

    All equalities work both from left to right and vice versa, from right to left. You can write:

    ab+ac = a(b+c)

    ab+ac+ad+.... = a(b+c+d+.....)

    That's the whole point of taking out common multiplier out of brackets.

    On the left side A - common multiplier for all terms. Multiplied by everything that exists). On the right is the most A is already located outside the brackets.

    Practical Application Let's look at the method using examples. At first the option is simple, even primitive.) But on this option I will note ( green) Very important points for any factorization.

    Factorize:

    ah+9x

    Which general does the multiplier appear in both terms? X, of course! We will put it out of brackets. Let's do this. We immediately write X outside the brackets:

    ax+9x=x(

    And in parentheses we write the result of division each term on this very X. In order:

    That's it. Of course, there is no need to describe it in such detail, this is done in the mind. But it is advisable to understand what’s what). We record in memory:

    We write the common factor outside the brackets. In parentheses we write the results of dividing all terms by this common factor. In order.

    So we have expanded the expression ah+9x by multipliers. Turned it into multiplying x by (a+9). I note that in the original expression there was also a multiplication, even two: a·x and 9·x. But it was not factorized! Because in addition to multiplication, this expression also contained addition, the “+” sign! And in expression x(a+9) There is nothing but multiplication!

    How so!? - I hear the indignant voice of the people - And in brackets!?)

    Yes, there is addition inside the parentheses. But the trick is that while the brackets are not opened, we consider them like one letter. And we do all the actions with brackets entirely, as with one letter. In this sense, in the expression x(a+9) There is nothing except multiplication. This is the whole point of factorization.

    By the way, is it possible to somehow check whether we did everything correctly? Easily! It’s enough to multiply back what you put out (x) by brackets and see if it worked original expression? If it works, everything is great!)

    x(a+9)=ax+9x

    It worked.)

    There are no problems in this primitive example. But if there are several terms, and even with different signs... In short, every third student messes up). Therefore:

    If necessary, check the factorization by inverse multiplication.

    Factorize:

    3ax+9x

    We are looking for a common factor. Well, everything is clear with X, it can be taken out. Is there more general factor? Yes! This is a three. You can write the expression like this:

    3ax+3 3x

    Here it is immediately clear that the common factor will be 3x. Here we take it out:

    3ax+3 3x=3x(a+3)

    Spread out.

    What happens if you take it out only x? Nothing special:

    3ax+9x=x(3a+9)

    This will also be a factorization. But in this fascinating process, it is customary to lay out everything to the limit while there is an opportunity. Here in brackets there is an opportunity to put out a three. It will turn out:

    3ax+9x=x(3a+9)=3x(a+3)

    The same thing, only with one extra action.) Remember:

    When taking the common factor out of brackets, we try to take out maximum common multiplier.

    Shall we continue the fun?)

    Factor the expression:

    3akh+9х-8а-24

    What will we take away? Three, X? Nope... You can't. I remind you that you can only take out general multiplier that is in all terms of the expression. That's why he general. There is no such multiplier here... What, you don’t have to expand it!? Well, yes, we were so happy... Meet:

    2. Grouping.

    Actually, it’s hard to name the group in an independent way factorization. It's more of a way to get out complex example.) We need to group the terms so that everything works out. This can only be shown by example. So, we have the expression:

    3akh+9х-8а-24

    It can be seen that there are some common letters and numbers. But... General there is no multiplier to be in all terms. Let's not lose heart and break the expression into pieces. Let's group. So that each piece has a common factor, there is something to take away. How do we break it? Yes, we just put parentheses.

    Let me remind you that parentheses can be placed anywhere and however you want. Just the essence of the example hasn't changed. For example, you can do this:

    3akh+9х-8а-24=(3ax+9x)-(8a+24)

    Please pay attention to the second brackets! They are preceded by a minus sign, and 8a And 24 turned positive! If, to check, we open the brackets back, the signs will change, and we get original expression. Those. the essence of the expression from the brackets has not changed.

    But if you just inserted parentheses without taking into account the change of sign, for example, like this:

    3akh+9х-8а-24=(3ax+9x) -(8a-24 )

    it would be a mistake. On the right - already other expression. Open the brackets and everything will become visible. You don’t have to decide further, yes...)

    But let's return to factorization. Let's look at the first brackets (3ax+9x) and we think, is there anything we can take out? Well, we solved this example above, we can take it 3x:

    (3ax+9x)=3x(a+3)

    Let's study the second brackets, we can add an eight there:

    (8a+24)=8(a+3)

    Our entire expression will be:

    (3ax+9x)-(8a+24)=3x(a+3)-8(a+3)

    Factored? No. The result of decomposition should be only multiplication but with us the minus sign spoils everything. But... Both terms have a common factor! This (a+3). It was not for nothing that I said that the entire brackets are, as it were, one letter. This means that these brackets can be taken out of brackets. Yes, that's exactly what it sounds like.)

    We do as described above. We write the common factor (a+3), in the second brackets we write the results of dividing the terms by (a+3):

    3x(a+3)-8(a+3)=(a+3)(3x-8)

    All! There is nothing on the right except multiplication! This means that factorization has been completed successfully!) Here it is:

    3ax+9x-8a-24=(a+3)(3x-8)

    Let us briefly repeat the essence of the group.

    If the expression does not general multiplier for everyone terms, we break the expression into brackets so that inside the brackets the common factor was. We take it out and see what happens. If you are lucky and there are absolutely identical expressions left in the brackets, we move these brackets out of brackets.

    I will add that grouping is a creative process). It doesn't always work out the first time. It's OK. Sometimes you have to swap terms and consider different options groups until a successful one is found. The main thing here is not to lose heart!)

    Examples.

    Now, having enriched yourself with knowledge, you can solve tricky examples.) At the beginning of the lesson there were three of these...

    Simplify:

    In essence, we have already solved this example. Unbeknownst to ourselves.) I remind you: if we are given a terrible fraction, we try to factor the numerator and denominator. Other simplification options just no.

    Well, the denominator here is not expanded, but the numerator... We have already expanded the numerator during the lesson! Like this:

    3ax+9x-8a-24=(a+3)(3x-8)

    We write the result of the expansion into the numerator of the fraction:

    According to the rule of reducing fractions (the main property of a fraction), we can divide (at the same time!) the numerator and denominator by the same number, or expression. Fraction from this doesn't change. So we divide the numerator and denominator by the expression (3x-8). And here and there we will get ones. The final result of the simplification:

    I would like to especially emphasize: reducing a fraction is possible if and only if in the numerator and denominator, in addition to multiplying expressions there is nothing. That is why the transformation of the sum (difference) into multiplication so important for simplification. Of course, if the expressions different, then nothing will be reduced. It will happen. But factorization gives a chance. This chance without decomposition is simply not there.

    Example with equation:

    Solve the equation:

    x 5 - x 4 = 0

    We take out the common factor x 4 out of brackets. We get:

    x 4 (x-1)=0

    We realize that the product of factors is equal to zero then and only then, when any of them is zero. If in doubt, find me a couple of non-zero numbers that, when multiplied, will give zero.) So we write, first the first factor:

    With such an equality, the second factor does not concern us. Anyone can be, but in the end it will still be zero. What number to the fourth power does zero give? Only zero! And no other... Therefore:

    We figured out the first factor and found one root. Let's look at the second factor. Now we don’t care about the first factor anymore.):

    Here we found a solution: x 1 = 0; x 2 = 1. Any of these roots fits our equation.

    Very important note. Please note that we solved the equation piece by piece! Each factor was equal to zero, regardless of other factors. By the way, if in such an equation there are not two factors, like ours, but three, five, as many as you like, we will solve exactly the same. Piece by piece. For example:

    (x-1)(x+5)(x-3)(x+2)=0

    Anyone who opens the brackets and multiplies everything will be stuck on this equation forever.) A correct student will immediately see that there is nothing on the left except multiplication, and zero on the right. And he will begin (in his mind!) to equate all brackets in order to zero. And he will receive (in 10 seconds!) right decision: x 1 = 1; x 2 = -5; x 3 = 3; x 4 = -2.

    Great, right?) Such an elegant solution is possible if left side equations factorized. Got the hint?)

    Well, last example, for older children):

    Solve the equation:

    It’s somewhat similar to the previous one, don’t you think?) Of course. It's time to remember that in seventh grade algebra, sines, logarithms, and anything else can be hidden under the letters! Factoring works throughout mathematics.

    We take out the common factor lg 4 x out of brackets. We get:

    log 4 x=0

    This is one root. Let's look at the second factor.

    Here is the final answer: x 1 = 1; x 2 = 10.

    I hope you've realized the power of factoring in simplifying fractions and solving equations.)

    In this lesson we learned about common factoring and grouping. It remains to deal with the formulas for abbreviated multiplication and the quadratic trinomial.

    If you like this site...

    By the way, I have a couple more interesting sites for you.)

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    Given online calculator decomposes numbers into prime factors by enumerating prime factors. If the number is large, then for ease of presentation, use a digit separator.

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    Factoring a number into prime factors - theory, algorithm, examples and solutions

    One of the simplest ways to factor a number is to check whether the number is divisible by 2, 3, 5,... etc., i.e. check if a number is divisible by a series of prime numbers. If the number n is not divisible by any prime number up to , then this number is prime, because if the number is composite, then it has at least two factors and both of them cannot be greater than .

    Let's imagine the number decomposition algorithm n into prime factors. Let's prepare a table of prime numbers in advance s=. Let us denote a series of prime numbers by p 1 , p 2 , p 3 , ...

    Algorithm for decomposing a number into prime factors:

    Example 1. Factor the number 153 into prime factors.

    Solution. It is enough for us to have a table of prime numbers up to , i.e. 2, 3, 5, 7, 11.

    Divide 153 by 2. 153 is not divisible by 2 without a remainder. Next, divide 153 by the next element of the table of prime numbers, i.e. at 3. 153:3=51. Fill out the table:

    Next, we check whether the number 17 is divisible by 3. The number 17 is not divisible by 3. It is not divisible by the numbers 5, 7, 11. The next divisor is greater . Therefore, 17 is a prime number that is divisible only by itself: 17:17=1. The procedure has stopped. fill out the table:

    We choose those divisors by which the numbers 153, 51, 17 are divided without a remainder, i.e. all numbers from right side tables. These are the divisors 3, 3, 17. Now the number 153 can be represented as a product of prime numbers: 153=3·3·17.

    Example 2. Factor the number 137 into prime factors.

    Solution. We calculate . This means we need to check the divisibility of the number 137 by prime numbers up to 11: 2,3,5,7,11. By dividing the number 137 by these numbers one by one, we find out that the number 137 is not divisible by any of the numbers 2,3,5,7,11. Therefore 137 is a prime number.

    Any composite number can be represented as a product of its prime divisors:

    28 = 2 2 7

    The right-hand sides of the resulting equalities are called prime factorization numbers 15 and 28.

    To factor a given composite number into prime factors means to represent this number as a product of its prime factors.

    The decomposition of a given number into prime factors is performed as follows:

    1. First you need to select the smallest prime number from the table of prime numbers that divides the given composite number without a remainder, and perform the division.
    2. Next, you need to again select the smallest prime number by which the already obtained quotient will be divided without a remainder.
    3. The second action is repeated until one is obtained in the quotient.

    As an example, let's factorize the number 940 into prime factors. Find the smallest prime number that divides 940. This number is 2:

    Now we select the smallest prime number that is divisible by 470. This number is again 2:

    The smallest prime number that is divisible by 235 is 5:

    The number 47 is prime, meaning the smallest prime number, which divides 47, will be this number itself:

    Thus, we get the number 940, factored into prime factors:

    940 = 2 470 = 2 2 235 = 2 2 5 47

    If the decomposition of a number into prime factors resulted in several identical factors, then for brevity, they can be written in the form of a power:

    940 = 2 2 5 47

    It is most convenient to write decomposition into prime factors as follows: first we write down this composite number and draw a vertical line to the right of it:

    To the right of the line we write the smallest prime divisor by which the given composite number is divided:

    We perform the division and write the resulting quotient under the dividend:

    We act with the quotient in the same way as with the given composite number, i.e., we select the smallest prime number by which it is divisible without a remainder and perform the division. And we repeat this until we get a unit in the quotient:

    Please note that sometimes it can be quite difficult to factor a number into prime factors, since when factoring we may encounter a large number, which is difficult to immediately determine whether it is simple or compound. And if it is composite, then it is not always easy to find its smallest prime divisor.

    Let’s try, for example, to factorize the number 5106 into prime factors:

    Having reached the quotient 851, it is difficult to immediately determine its smallest divisor. We turn to the table of prime numbers. If there is a number in it that puts us in difficulty, then it is divisible only by itself and one. The number 851 is not in the table of prime numbers, which means it is composite. All that remains is to divide it into prime numbers by sequential enumeration: 3, 7, 11, 13, ..., and so on until we find a suitable prime divisor. By brute force we find that 851 is divisible by the number 23.


    In this article you will find all necessary information answering the question how to factor a number into prime factors. First given general idea about the decomposition of a number into prime factors, examples of decompositions are given. The following shows the canonical form of decomposing a number into prime factors. After this, an algorithm is given for decomposing arbitrary numbers into prime factors and examples of decomposing numbers using this algorithm are given. Also considered alternative ways, which allow you to quickly factor small integers into prime factors using divisibility tests and multiplication tables.

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

    First, let's look at what prime factors are.

    It is clear that since the word “factors” is present in this phrase, then there is a product of some numbers, and the qualifying word “simple” means that each factor is a prime number. For example, in a product of the form 2·7·7·23 there are four prime factors: 2, 7, 7 and 23.

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

    This means that this number must be represented as a product of prime factors, and the value of this product must be equal to original number. As an example, consider the product of three prime numbers 2, 3 and 5, it is equal to 30, thus the decomposition of the number 30 into prime factors is 2·3·5. Usually the decomposition of a number into prime factors is written as an equality; in our example it will be like this: 30=2·3·5. We emphasize separately that prime factors in the expansion can be repeated. This clearly illustrates next example: 144=2·2·2·2·3·3 . But a representation of the form 45=3·15 is not a decomposition into prime factors, since the number 15 is a composite number.

    The following question arises: “What numbers can be decomposed into prime factors?”

    In search of an answer to it, we present the following reasoning. Prime numbers, by definition, are among those greater than one. Taking into account this fact and , it can be argued that the product of several prime factors is an integer positive number, exceeding one. Therefore, factorization takes place only for positive integers that are greater than 1.

    But can all integers greater than one be factored into prime factors?

    It is clear that it is not possible to factor simple integers into prime factors. This is explained by the fact that prime numbers have only two positive divisors - one and itself, so they cannot be represented as a product of two or more prime numbers. If the integer z could be represented as the product of prime numbers a and b, then the concept of divisibility would allow us to conclude that z is divisible by both a and b, which is impossible due to the simplicity of the number z. However, they believe that any prime number is itself a decomposition.

    What about composite numbers? Are composite numbers decomposed into prime factors, and are all composite numbers subject to such decomposition? The fundamental theorem of arithmetic gives an affirmative answer to a number of these questions. The basic theorem of arithmetic states that any integer a that is greater than 1 can be decomposed into the product of prime factors p 1, p 2, ..., p n, and the decomposition has the form a = p 1 · p 2 ·… · p n, and this the expansion is unique, if you do not take into account the order of the factors

    Canonical factorization of a number into prime factors

    In the expansion of a number, prime factors can be repeated. Repeating prime factors can be written more compactly using . Let in the decomposition of a number the prime factor p 1 occur s 1 times, the prime factor p 2 – s 2 times, and so on, p n – s n times. Then the prime factorization of the number a can be written as a=p 1 s 1 ·p 2 s 2 ·…·p n s n. This form of recording is the so-called canonical factorization of a number into prime factors.

    Let us give an example of the canonical decomposition of a number into prime factors. Let us know the decomposition 609 840=2 2 2 2 3 3 5 7 11 11, its canonical notation has the form 609 840=2 4 3 2 5 7 11 2.

    The canonical factorization of a number into prime factors allows you to find all the divisors of the number and the number of divisors of the number.

    Algorithm for factoring a number into prime factors

    To successfully cope with the task of decomposing a number into prime factors, you need to have a very good knowledge of the information in the article prime and composite numbers.

    The essence of the process of decomposing a positive integer number a that exceeds one is clear from the proof of the fundamental theorem of arithmetic. The point is to sequentially find the smallest prime divisors p 1, p 2, ..., p n of the numbers a, a 1, a 2, ..., a n-1, which allows us to obtain a series of equalities 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 . When we get a n =1, then the equality a=p 1 ·p 2 ·…·p n will give us the desired decomposition of the number a into prime factors. It should also be noted here that p 1 ≤p 2 ≤p 3 ≤…≤p n.

    It remains to figure out how to find the smallest prime factors at each step, and we will have an algorithm for decomposing a number into prime factors. A table of prime numbers will help us find prime factors. Let us show how to use it to obtain the smallest prime divisor of the number z.

    We sequentially take prime numbers from the table of prime numbers (2, 3, 5, 7, 11, and so on) and divide the given number z by them. The first prime number by which z is evenly divided will be its smallest prime divisor. If the number z is prime, then its smallest prime divisor will be the number z itself. It should be recalled here that if z is not a prime number, then its smallest prime divisor does not exceed the number , where is from z. Thus, if among the prime numbers not exceeding , there was not a single divisor of the number z, then we can conclude that z is a prime number (more about this is written in the theory section under the heading This number is prime or composite).

    As an example, we will show how to find the smallest prime divisor of the number 87. Let's take the number 2. Divide 87 by 2, we get 87:2=43 (remaining 1) (if necessary, see article). That is, when dividing 87 by 2, the remainder is 1, so 2 is not a divisor of the number 87. We take the next prime number from the prime numbers table, this is the number 3. Divide 87 by 3, we get 87:3=29. Thus, 87 is divisible by 3, therefore, the number 3 is the smallest prime divisor of the number 87.

    Note that in the general case, to factorize the number a into prime factors, we need a table of prime numbers up to a number not less than . We will have to refer to this table at every step, so we need to have it at hand. For example, to factorize the number 95 into prime factors, we will only need a table of prime numbers up to 10 (since 10 is greater than ). And to decompose the number 846,653, you will already need a table of prime numbers up to 1,000 (since 1,000 is greater than ).

    We now have enough information to write down algorithm for factoring a number into prime factors. The algorithm for decomposing the number a is as follows:

    • Sequentially sorting through the numbers from the table of prime numbers, we find the smallest prime divisor p 1 of the number a, after which we calculate a 1 =a:p 1. If a 1 =1, then the number a is prime, and it itself is its decomposition into prime factors. If a 1 is not equal to 1, then we have a=p 1 ·a 1 and move on to the next step.
    • We find the smallest prime divisor p 2 of the number a 1 , to do this we sequentially sort through the numbers from the table of prime numbers, starting with p 1 , and then calculate a 2 =a 1:p 2 . If a 2 =1, then the required decomposition of the number a into prime factors has the form a=p 1 ·p 2. If a 2 is not equal to 1, then we have a=p 1 ·p 2 ·a 2 and move on to the next step.
    • Going through the numbers from the table of prime numbers, starting with p 2, we find the smallest prime divisor p 3 of the number a 2, after which we calculate a 3 =a 2:p 3. If a 3 =1, then the required decomposition of the number a into prime factors has the form a=p 1 ·p 2 ·p 3. If a 3 is not equal to 1, then we have a=p 1 ·p 2 ·p 3 ·a 3 and move on to the next step.
    • We find the smallest prime divisor p n of the number a n-1 by sorting through the prime numbers, starting with p n-1, as well as a n =a n-1:p n, and a n is equal to 1. This step is the last step of the algorithm; here we obtain the required decomposition of the number a into prime factors: a=p 1 ·p 2 ·…·p n.

    For clarity, all the results obtained at each step of the algorithm for decomposing a number into prime factors are presented in the form of the following table, in which the numbers a, a 1, a 2, ..., a n are written sequentially in a column to the left of the vertical line, and to the right of the line - the corresponding smallest prime divisors p 1, p 2, ..., p n.

    All that remains is to consider a few examples of the application of the resulting algorithm for decomposing numbers into prime factors.

    Examples of prime factorization

    Now we will look in detail examples of factoring numbers into prime factors. When decomposing, we will use the algorithm from the previous paragraph. Let's start with simple cases, and we will gradually complicate them in order to encounter all the possible nuances that arise when decomposing numbers into simple factors.

    Example.

    Factor the number 78 into its prime factors.

    Solution.

    We begin the search for the first smallest prime divisor p 1 of the number a=78. To do this, we begin to sequentially sort through prime numbers from the table of prime numbers. We take the number 2 and divide 78 by it, we get 78:2=39. The number 78 is divided by 2 without a remainder, so p 1 =2 is the first found prime divisor of the number 78. In this case, a 1 =a:p 1 =78:2=39. So we come to the equality a=p 1 ·a 1 having the form 78=2·39. Obviously, a 1 =39 is different from 1, so we move on to the second step of the algorithm.

    Now we are looking for the smallest prime divisor p 2 of the number a 1 =39. We begin enumerating numbers from the table of prime numbers, starting with p 1 =2. Divide 39 by 2, we get 39:2=19 (remaining 1). Since 39 is not evenly divisible by 2, then 2 is not its divisor. Then we take the next number from the table of prime numbers (number 3) and divide 39 by it, we get 39:3=13. Therefore, p 2 =3 is the smallest prime divisor of the number 39, while a 2 =a 1:p 2 =39:3=13. We have the equality a=p 1 ·p 2 ·a 2 in the form 78=2·3·13. Since a 2 =13 is different from 1, we move on to the next step of the algorithm.

    Here we need to find the smallest prime divisor of the number a 2 =13. In search of the smallest prime divisor p 3 of the number 13, we will sequentially sort through the numbers from the table of prime numbers, starting with p 2 =3. The number 13 is not divisible by 3, since 13:3=4 (rest. 1), also 13 is not divisible by 5, 7 and 11, since 13:5=2 (rest. 3), 13:7=1 (rest. 6) and 13:11=1 (rest. 2). The next prime number is 13, and 13 is divisible by it without a remainder, therefore, the smallest prime divisor p 3 of 13 is the number 13 itself, and a 3 =a 2:p 3 =13:13=1. Since a 3 =1, this step of the algorithm is the last, and the required decomposition of the number 78 into prime factors has the form 78=2·3·13 (a=p 1 ·p 2 ·p 3 ).

    Answer:

    78=2·3·13.

    Example.

    Express the number 83,006 as a product of prime factors.

    Solution.

    At the first step of the algorithm for decomposing a number into prime factors, we find p 1 =2 and a 1 =a:p 1 =83,006:2=41,503, from which 83,006=2·41,503.

    At the second step, we find out that 2, 3 and 5 are not prime divisors of the number a 1 =41,503, but the number 7 is, since 41,503:7=5,929. We have p 2 =7, a 2 =a 1:p 2 =41,503:7=5,929. Thus, 83,006=2 7 5 929.

    The smallest prime divisor of the number a 2 =5 929 is the number 7, since 5 929:7 = 847. Thus, p 3 =7, a 3 =a 2:p 3 =5 929:7 = 847, from which 83 006 = 2·7·7·847.

    Next we find that the smallest prime divisor p 4 of the number a 3 =847 is equal to 7. Then a 4 =a 3:p 4 =847:7=121, so 83 006=2·7·7·7·121.

    Now we find the smallest prime divisor of the number a 4 =121, it is the number p 5 =11 (since 121 is divisible by 11 and not divisible by 7). Then a 5 =a 4:p 5 =121:11=11, and 83 006=2·7·7·7·11·11.

    Finally, the smallest prime divisor of the number a 5 =11 is the number p 6 =11. Then a 6 =a 5:p 6 =11:11=1. Since a 6 =1, this step of the algorithm for decomposing a number into prime factors is the last, and the desired decomposition has the form 83 006 = 2·7·7·7·11·11.

    The result obtained can be written as the canonical decomposition of the number into prime factors 83 006 = 2·7 3 ·11 2.

    Answer:

    83 006=2 7 7 7 11 11=2 7 3 11 2 991 is a prime number. Indeed, it does not have a single prime divisor not exceeding ( can be roughly estimated as , since it is obvious that 991<40 2 ), то есть, наименьшим делителем числа 991 является оно само. Тогда p 3 =991 и a 3 =a 2:p 3 =991:991=1 . Следовательно, искомое разложение числа 897 924 289 на простые множители имеет вид 897 924 289=937·967·991 .

    Answer:

    897 924 289 = 937 967 991 .

    Using divisibility tests for prime factorization

    In simple cases, you can decompose a number into prime factors without using the decomposition algorithm from the first paragraph of this article. If the numbers are not large, then to decompose them into prime factors it is often enough to know the signs of divisibility. Let's give examples for clarification.

    For example, we need to factor the number 10 into prime factors. From the multiplication table we know that 2·5=10, and the numbers 2 and 5 are obviously prime, so the prime factorization of the number 10 looks like 10=2·5.

    Another example. Using the multiplication table, we will factor the number 48 into prime factors. We know that six is ​​eight - forty-eight, that is, 48 ​​= 6·8. However, neither 6 nor 8 are prime numbers. But we know that twice three is six, and twice four is eight, that is, 6=2·3 and 8=2·4. Then 48=6·8=2·3·2·4. It remains to remember that two times two is four, then we get the desired decomposition into prime factors 48 = 2·3·2·2·2. Let's write this expansion in canonical form: 48=2 4 ·3.

    But when factoring the number 3,400 into prime factors, you can use the divisibility criteria. The signs of divisibility by 10, 100 allow us to state that 3,400 is divisible by 100, with 3,400=34·100, and 100 is divisible by 10, with 100=10·10, therefore, 3,400=34·10·10. And based on the test of divisibility by 2, we can say that each of the factors 34, 10 and 10 is divisible by 2, we get 3 400=34 10 10=2 17 2 5 2 5. All factors in the resulting expansion are simple, so this expansion is the desired one. All that remains is to rearrange the factors so that they go in ascending order: 3 400 = 2·2·2·5·5·17. Let us also write down the canonical decomposition of this number into prime factors: 3 400 = 2 3 ·5 2 ·17.

    When decomposing a given number into prime factors, you can use in turn both the signs of divisibility and the multiplication table. Let's imagine the number 75 as a product of prime factors. The test of divisibility by 5 allows us to state that 75 is divisible by 5, and we obtain that 75 = 5·15. And from the multiplication table we know that 15=3·5, therefore, 75=5·3·5. This is the required decomposition of the number 75 into prime factors.

    References.

    • Vilenkin N.Ya. and others. Mathematics. 6th grade: textbook for general education institutions.
    • Vinogradov I.M. Fundamentals of number theory.
    • Mikhelovich Sh.H. Number theory.
    • Kulikov L.Ya. and others. Collection of problems in algebra and number theory: Textbook for students of physics and mathematics. specialties of pedagogical institutes.