Mathematical expressions and problems require a lot of additional knowledge. NOC is one of the main ones, especially often used in The subject is studied in high school, while it is not particularly difficult to understand material, it is not difficult for a person familiar with the degrees and multiplication table to select the necessary numbers and find the result.

Definition

The common multiple is a number that can completely divide into two numbers at the same time (a and b). Most often, this number is obtained by multiplying the original numbers a and b. The number must be divided immediately into both numbers, without deviations.

NOC is the short name used for designation, collected from the first letters.

Ways to get a number

The method of multiplying numbers is not always suitable for finding NOCs; it is much better suited for simple single-digit or two-digit numbers. it is customary to factor, the larger the number, the more factors there will be.

Example No. 1

For the simplest example, schools usually take simple, single, or double-digit numbers. For example, you need to solve the following task, find the least common multiple of the numbers 7 and 3, the solution is quite simple, just multiply them. As a result, there is the number 21, a smaller number is simply not there.

Example No. 2

The second option is much more complicated. The numbers 300 and 1260 are given; finding the NOC is mandatory. The following actions are supposed to solve the task:

Decomposition of the first and second numbers into simple factors. 300 \u003d 2 2 * 3 * 5 2; 1260 \u003d 2 2 * 3 2 * 5 * 7. The first stage is completed.

The second stage involves working with data already received. Each of the numbers received is required to participate in the calculation of the final result. For each factor, the largest number of occurrences is taken from the composition of the original numbers. NOC is a total number, so the factors of the numbers should be repeated in it all to one, even those that are present in one instance. Both initial numbers have in their composition the numbers 2, 3 and 5, in different degrees, 7 is in only one case.

To calculate the final result, it is necessary to take each number in the largest degree presented, in the equation. It remains only to multiply and get an answer, with proper completion the task fits into two actions without explanation:

1) 300 = 2 2 * 3 * 5 2 ; 1260 = 2 2 * 3 2 *5 *7.

2) NOC \u003d 6300.

That's the whole problem, if you try to calculate the desired number by multiplying, then the answer will definitely not be correct, since 300 * 1260 \u003d 378,000.

Verification:

6300/300 \u003d 21 - true;

6300/1260 \u003d 5 - true.

The correctness of the result is determined by checking - dividing the NOC by both initial numbers, if the number is integer in both cases, then the answer is correct.

What does NOC mean in mathematics?

As you know, in mathematics there is not a single useless function, this is no exception. The most common purpose of this number is to bring fractions to a common denominator. What is usually studied in grades 5-6 of high school. It is also additionally a common divisor for all multiple numbers, if such conditions are in the problem. A similar expression can find a multiple not only to two numbers, but also to a much larger number - three, five, and so on. The more numbers, the more actions in the task, but the complexity of this does not increase.

For example, given the numbers 250, 600 and 1500, you need to find their total NOC:

1) 250 \u003d 25 * 10 \u003d 5 2 * 5 * 2 \u003d 5 3 * 2 - this example describes in detail the factorization, without reduction.

2) 600 = 60 * 10 = 3 * 2 3 *5 2 ;

3) 1500 = 15 * 100 = 33 * 5 3 *2 2 ;

In order to compose an expression, it is necessary to mention all the factors, in this case, 2, 5, 3 are given - for all these numbers it is necessary to determine the maximum degree.

Attention: all factors must be brought to full simplification, if possible, expanding to the level of single-valued ones.

Verification:

1) 3000/250 \u003d 12 - true;

2) 3000/600 \u003d 5 - true;

3) 3000/1500 \u003d 2 - true.

This method does not require any tricks or abilities of the genius level, everything is simple and clear.

Another way

In mathematics, a lot is connected, a lot can be solved in two or more ways, the same goes for finding the least common multiple, NOC. The following method can be used in the case of simple two-digit and single-digit numbers. A table is compiled, in which the multiplied vertically is entered, the multiplier horizontally, and the product is indicated in the intersecting cells of the column. You can reflect the table by means of a line, take a number and write in a row the results of multiplying this number by integers, from 1 to infinity, sometimes 3-5 points are enough, the second and subsequent numbers undergo the same computational process. Everything happens until a common multiple is found.

Given the numbers 30, 35, 42, it is necessary to find the NOC linking all the numbers:

1) Multiples of 30: 60, 90, 120, 150, 180, 210, 250, etc.

2) Multiples of 35: 70, 105, 140, 175, 210, 245, etc.

3) Multiples of 42: 84, 126, 168, 210, 252, etc.

It is noticeable that all numbers are quite different, the only common number among them is 210, so it will be the NOC. Among the processes associated with this calculation, there is also the largest common divisor, calculated by similar principles and often found in neighboring problems. The difference is small, but significant enough, the NOC involves the calculation of a number that is divided by all the original values, and the GCD involves the calculation of the largest value by which the original numbers are divided.

How to find the NOC (smallest common multiple)

   The common multiple for two integers is an integer that is divisible without a remainder by both given numbers.

The smallest common multiple for two integers is the smallest of all integers that is divisible completely and without remainder by both given numbers.

Method 1. You can find the NOCs, in turn, for each of the given numbers, writing out in ascending order all the numbers that are obtained by multiplying them by 1, 2, 3, 4, and so on.

Example   for numbers 6 and 9.
   We multiply the number 6, sequentially, by 1, 2, 3, 4, 5.
   We get: 6, 12, 18 , 24, 30
   We multiply the number 9, sequentially, by 1, 2, 3, 4, 5.
   We get: 9, 18 , 27, 36, 45
   As you can see, the NOC for the numbers 6 and 9 will be equal to 18.

This method is convenient when both numbers are small and it is easy to multiply them by a sequence of integers. However, there are times when you need to find the NOC for two-digit or three-digit numbers, as well as when the initial numbers are three or even more.

Method 2. You can find the NOC by decomposing the original numbers into prime factors.
After the expansion, it is necessary to cross out the same numbers from the resulting series of prime factors. The remaining numbers of the first number will be a factor for the second, and the remaining numbers of the second will be a factor for the first.

Examplefor the numbers 75 and 60.
   The least common multiple of numbers 75 and 60 can be found without writing out consecutive multiples of these numbers. To do this, we decompose 75 and 60 into simple factors:
75 = 3 * 5   * 5, and
60 = 2 * 2 * 3 * 5 .
   As you can see, factors 3 and 5 are found on both lines. Mentally cross them out.
   We write out the remaining factors included in the expansion of each of these numbers. When expanding the number 75, we have the number 5 left, and when expanding the number 60, we have 2 * 2
   So, to determine the NOC for the numbers 75 and 60, we need to multiply the remaining numbers from the decomposition of 75 (this is 5) by 60, and the numbers remaining from the decomposition of the number 60 (this is 2 * 2) multiply by 75. That is, for ease of understanding , we say that we multiply "crosswise".
75 * 2 * 2 = 300
60 * 5 = 300
   So we found the NOC for the numbers 60 and 75. This is the number 300.

Example. Define NOC for numbers 12, 16, 24
   In this case, our actions will be somewhat more complicated. But first, as always, we decompose all the numbers into prime factors
12 = 2 * 2 * 3
16 = 2 * 2 * 2 * 2
24 = 2 * 2 * 2 * 3
   In order to correctly determine the NOC, we select the smallest of all numbers (this is the number 12) and sequentially go through its factors, crossing them out if at least one of the other rows of numbers contains the same, not crossed out multiplier.

Step 1 . We see that 2 * 2 are found in all rows of numbers. Cross them out.
12 = 2 * 2 * 3
16 = 2 * 2 * 2 * 2
24 = 2 * 2 * 2 * 3

Step 2. In the prime factors of the number 12, only the number 3 remains. But it is present in the prime factors of the number 24. Cross out the number 3 from both series, and no action is supposed for the number 16.
12 = 2 * 2 * 3
16 = 2 * 2 * 2 * 2
24 = 2 * 2 * 2 * 3

As we see, in the expansion of the number 12, we “crossed out” all the numbers. Means finding NOC completed. It remains only to calculate its value.
   For the number 12, we take the remaining factors from the number 16 (the nearest ascending)
12 * 2 * 2 = 48
   This is the NOC

As you can see, in this case, finding the NOC was somewhat more difficult, but when you need to find it for three or more numbers, this method allows you to do this faster. However, both methods of finding NOCs are correct.

Pupils are given many tasks in mathematics. Among them, problems with such a formulation are very common: there are two meanings. How to find the smallest common multiple for given numbers? It is necessary to be able to perform such tasks, since the acquired skills are used to work with fractions with different denominators. In the article we will analyze how to find the NOC and the basic concepts.

Before you find the answer to the question of how to find the NOC, you need to decide on the term multiple. Most often, the wording of this concept is as follows: a multiple of a certain value of A is called a natural number that will be divisible by A. Without a remainder, 8, 12, 16, 20, and so on, up to the necessary limit, will be multiples of 4.

Moreover, the number of divisors for a particular value can be limited, and infinitely many multiples. Also there is the same value for natural values. This is an indicator that is divided into them without a trace. Having dealt with the concept of the smallest value for certain indicators, we will move on to how to find it.

Find the NOC

The smallest multiple of two or more indicators is the smallest natural number that is completely divided by all the numbers indicated.

There are several ways to find such a value., consider the following methods:

1. If the numbers are small, then write down on the line all dividing into it. Keep doing this until you find a common among them. In the record they are denoted by the letter K. For example, for 4 and 3, the smallest multiple is 12.
2. If it is large or if you want to find a multiple for 3 or more values, then you should use another technique that involves decomposing numbers into prime factors. First, lay out the largest of the specified, then all the rest. Each of them has its own number of factors. As an example, we decompose 20 (2 * 2 * 5) and 50 (5 * 5 * 2). For the smaller of them, emphasize the factors and add to the largest. The result is 100, which will be the smallest common multiple for the above numbers.
3. When finding 3 numbers (16, 24 and 36), the principles are the same as for the other two. Let us expand each of them: 16 \u003d 2 * 2 * 2 * 2, 24 \u003d 2 * 2 * 2 * 3, 36 \u003d 2 * 2 * 3 * 3. Only two deuces from the decomposition of the number 16 were not included in the decomposition of the greatest. We add them and we get 144, which is the smallest result for the numerical values \u200b\u200bindicated earlier.

Now we know what is the general methodology for finding the smallest value for two, three or more values.   However, there are private methods.helping to look for NOCs if previous ones don't help.

How to find GCD and NOC.

Private ways to find

As for any mathematical section, there are special cases of finding NOCs that help in specific situations:

• if one of the numbers is divided into others without a remainder, then the lowest multiple of these numbers is equal to it (LCL 60 and 15 is 15);
• mutually prime numbers do not have common prime divisors. Their smallest value is equal to the product of these numbers. Thus, for numbers 7 and 8, this will be 56;
• the same rule also works for other cases, including special cases, which can be read in the specialized literature. This also includes cases of decomposition of composite numbers, which are the topic of individual articles and even candidate dissertations.

Special cases are less common than standard examples. But thanks to them, you can learn to work with fractions of varying degrees of complexity. This is especially true for fractions.where there are different denominators.

Some examples

Let's look at a few examples, thanks to which we can understand the principle of finding the least multiple:

1. We find the NOC (35; 40). We decompose first 35 \u003d 5 * 7, then 40 \u003d 5 * 8. Add 8 to the smallest digit and get NOC 280.
2. NOC (45; 54). We decompose each of them: 45 \u003d 3 * 3 * 5 and 54 \u003d 3 * 3 * 6. Add the number 6 to 45. We get the NOC equal to 270.
3. Well, the last example. There are 5 and 4. There are no simple multiples for them, so the least common multiple in this case will be their product, equal to 20.

Thanks to examples, you can understand how the NOC is located, what are the nuances and what is the meaning of such manipulations.

Finds the NOC is much easier than it might seem initially. For this, both simple decomposition and multiplication of simple values \u200b\u200bby each other are used. The ability to work with this section of mathematics helps in the further study of mathematical topics, especially fractions of varying degrees of complexity.

Do not forget to periodically solve examples using various methods, this develops a logical apparatus and allows you to remember numerous terms. Learn the methods of finding such an indicator and you can work well with the rest of the mathematical sections. Have fun learning math!

Video

This video will help you understand and remember how to find the smallest common multiple.

We continue the discussion about the least common multiple that we started in the section “NOC - the smallest common multiple, definition, examples”. In this topic, we will look at ways to find NOCs for three or more numbers; we will examine the question of how to find an NOC of a negative number.

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Calculation of the smallest common multiple (NLC) through GCD

We have already established the connection of the least common multiple with the largest common divisor. Now we will learn to determine the NOC through the GCD. First, let's figure out how to do this for positive numbers.

Definition 1

The smallest common multiple can be found using the largest common factor by the formula NOC (a, b) \u003d a · b: GCD (a, b).

Example 1

You need to find the NOC numbers 126 and 70.

Decision

We take a \u003d 126, b \u003d 70. We substitute the values \u200b\u200bin the formula for calculating the least common multiple through the largest common divisor of the LCL (a, b) \u003d a · b: GCD (a, b).

Find the GCD of the numbers 70 and 126. For this we need the Euclidean algorithm: 126 \u003d 70 · 1 + 56, 70 \u003d 56 · 1 + 14, 56 \u003d 14 · 4, therefore, GCD (126 , 70) = 14 .

We calculate the NOC: NOC (126, 70) \u003d 12670: GCD (126, 70) \u003d 12670: 14 \u003d 630.

Answer:   NOC (126, 70) \u003d 630.

Example 2

Find the knock of numbers 68 and 34.

Decision

NOD in this case, it’s not difficult to neutralize, since 68 is divided by 34. We calculate the smallest common multiple by the formula: NOC (68, 34) \u003d 68 · 34: GCD (68, 34) \u003d 68 · 34: 34 \u003d 68.

Answer:   NOC (68, 34) \u003d 68.

In this example, we used the rule of finding the least common multiple for positive integers a and b: if the first number is divisible by the second, then the LCM of these numbers will be equal to the first number.

Finding NOCs by factoring numbers into prime factors

Now let's look at a way to find the NOC, which is based on the decomposition of numbers into prime factors.

Definition 2

To find the least common multiple, we need to perform a number of simple steps:

• we compose the product of all prime factors of numbers for which we need to find the NOC;
• exclude their obtained products all prime factors;
• the product obtained after eliminating the common prime factors will be equal to the NOC of these numbers.

This method of finding the smallest common multiple is based on the equality of the NOC (a, b) \u003d a · b: GCD (a, b). If you look at the formula, it becomes clear: the product of the numbers a and b is equal to the product of all the factors that participate in the expansion of these two numbers. Moreover, the GCD of two numbers is equal to the product of all prime factors that are simultaneously present in the factorizations of these two numbers.

Example 3

We have two numbers 75 and 210. We can factor them as follows:   75 \u003d 3 · 5 · 5   and   210 \u003d 2 · 3 · 5 · 7. If we compose the product of all factors of two source numbers, we get:   2 · 3 · 3 · 5 · 5 · 5 · 7.

If we exclude the common factors for both numbers, 3 and 5, we get a product of the following form:   2 · 3 · 5 · 5 · 7 \u003d 1050. This work will be our NOC for the numbers 75 and 210.

Example 4

Find the NOC of numbers 441   and 700 factoring both numbers into prime factors.

Decision

Find all the prime factors of the numbers given in the condition:

441 147 49 7 1 3 3 7 7

700 350 175 35 7 1 2 2 5 5 7

We get two chains of numbers: 441 \u003d 3 · 3 · 7 · 7 and 700 \u003d 2 · 2 · 5 · 5 · 7.

The product of all factors that participated in the decomposition of these numbers will look like:   2 · 2 · 3 · 3 · 5 · 5 · 7 · 7 · 7. Find the common factors. This is the number 7. We exclude it from the general work:   2 · 2 · 3 · 3 · 5 · 5 · 7 · 7. It turns out that the NOC   (441, 700) \u003d 2 · 2 · 3 · 3 · 5 · 5 · 7 · 7 \u003d 44 100.

Answer:   NOC (441, 700) \u003d 44 100.

Let us give one more formulation of the method for finding the LCL by decomposing numbers into prime factors.

Definition 3

Previously, we excluded common factors for both numbers from the total number of factors. Now we will do otherwise:

• factor both numbers into prime factors:
• add to the product of prime factors of the first number the missing factors of the second number;
• we get the product, which will be the desired NOC of two numbers.

Example 5

Let us return to the numbers 75 and 210, for which we have already looked for the NOC in one of the previous examples. We decompose them into prime factors:   75 \u003d 3 · 5 · 5   and   210 \u003d 2 · 3 · 5 · 7. To the product of factors 3, 5 and 5   numbers 75 add the missing factors 2   and 7   numbers 210. We get:   2 · 3 · 5 · 5 · 7.This is the NOC of the numbers 75 and 210.

Example 6

It is necessary to calculate the NOC numbers 84 and 648.

Decision

We decompose the numbers from the condition into prime factors:   84 \u003d 2 · 2 · 3 · 7   and   648 \u003d 2 · 2 · 2 · 3 · 3 · 3 · 3. Add to the product of factors 2, 2, 3 and 7   84 missing factors 2, 3, 3 and
3   numbers 648. We get the product   2 · 2 · 2 · 3 · 3 · 3 · 3 · 7 \u003d 4536.   This is the smallest common multiple of 84 and 648.

Answer:   NOC (84, 648) \u003d 4,536.

Finding NOCs of three or more numbers

Regardless of how many numbers we are dealing with, the algorithm of our actions will always be the same: we will successively find the NOC of two numbers. There is a theorem for this case.

Theorem 1

Suppose we have integers   a 1, a 2, ..., a k. NOC   m k   these numbers are found in the sequential calculation of m 2 \u003d NOC (a 1, a 2), m 3 \u003d NOC (m 2, a 3), ..., m k \u003d NOC (m k - 1, a k).

Now we will consider how the theorem can be applied to solve specific problems.

Example 7

It is necessary to calculate the smallest common multiple of the four numbers 140, 9, 54 and 250 .

Decision

We introduce the notation: a 1 \u003d 140, a 2 \u003d 9, a 3 \u003d 54, a 4 \u003d 250.

To begin with, we calculate m 2 \u003d NOC (a 1, a 2) \u003d NOC (140, 9). We apply the Euclidean algorithm to calculate the GCD of numbers 140 and 9: 140 \u003d 9 · 15 + 5, 9 \u003d 5 · 1 + 4, 5 \u003d 4 · 1 + 1, 4 \u003d 1 · 4. We get: GCD (140, 9) \u003d 1, NOC (140, 9) \u003d 140 · 9: GCD (140, 9) \u003d 140 · 9: 1 \u003d 1,260. Therefore, m 2 \u003d 1,260.

Now, by the same algorithm, we calculate m 3 \u003d NOC (m 2, a 3) \u003d NOC (1,260, 54). In the course of calculations, we obtain m 3 \u003d 3,780.

It remains for us to calculate m 4 \u003d NOC (m 3, a 4) \u003d NOC (3 780, 250). We follow the same algorithm. We get m 4 \u003d 94 500.

The NOC of four numbers from the conditions of the example is 94500.

Answer:   NOC (140, 9, 54, 250) \u003d 94 500.

As you can see, the calculations are simple, but rather time-consuming. To save time, you can go the other way.

Definition 4

We offer you the following algorithm of actions:

• we decompose all numbers into prime factors;
• add the missing factors from the product of the second number to the product of the factors of the first number;
• to the product obtained at the previous stage, add the missing factors of the third number, etc .;
• the resulting product will be the smallest common multiple of all numbers from the condition.

Example 8

It is necessary to find the NOC of the five numbers 84, 6, 48, 7, 143.

Decision

We decompose all five numbers into prime factors: 84 \u003d 2 · 2 · 3 · 7, 6 \u003d 2 · 3, 48 \u003d 2 · 2 · 2 · 2 · 3, 7, 143 \u003d 11 · 13. Prime numbers, which is the number 7, are not decomposed into prime factors. Such numbers coincide with their factorization.

Now take the product of prime factors 2, 2, 3 and 7 of 84 and add the missing factors of the second number to them. We decompose the number 6 into 2 and 3. These factors are already in the product of the first number. Therefore, we omit them.

We continue to add the missing factors. We pass to the number 48, from the product of prime factors of which we take 2 and 2. Then add the prime factor 7 of the fourth and the factors of 11 and 13 of the fifth. We get: 2 · 2 · 2 · 2 · 3 · 7 · 11 · 13 \u003d 48 048. This is the smallest common multiple of five source numbers.

Answer:   NOC (84, 6, 48, 7, 143) \u003d 48 048.

Finding the smallest common multiple of negative numbers

In order to find the smallest common multiple of negative numbers, these numbers must first be replaced with numbers with the opposite sign, and then carry out calculations using the above algorithms.

Example 9

NOC (54, - 34) \u003d NOC (54, 34), and NOC (- 622, - 46, - 54, - 888) \u003d NOC (622, 46, 54, 888).

Such actions are permissible due to the fact that if we accept that   a   and   - a   - opposite numbers,
  then many multiple numbers a   matches multiple plurals   - a.

Example 10

It is necessary to calculate the NOC of negative numbers − 145   and − 45 .

Decision

We will replace the numbers − 145   and − 45   on opposite numbers 145 and 45 . Now, according to the algorithm, we calculate the NOC (145, 45) \u003d 145 · 45: GCD (145, 45) \u003d 145 · 45: 5 \u003d 1 305, having previously determined the GCD using the Euclidean algorithm.

We get that the NOC of numbers is 145 and − 45   equally 1 305 .

Answer:   NOC (- 145, - 45) \u003d 1 305.

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The material presented below is a logical continuation of the theory from the article under the heading of the NOC - the least common multiple, definition, examples, the relationship between the NOC and GCD. Here we talk about finding the smallest common multiple (LCL), and we will pay special attention to solving examples. First, we show how the NOC of two numbers is calculated through the GCD of these numbers. Next, consider finding the smallest common multiple by decomposing the numbers into prime factors. After that, we will focus on finding NOCs of three or more numbers, and also pay attention to calculating NOCs of negative numbers.

Page navigation.

Calculation of the smallest common multiple (NLC) through GCD

One way to find the smallest common multiple is based on the relationship between the NOC and the GCD. The existing relationship between the NOC and the GCD allows us to calculate the smallest common multiple of two positive integers through the known largest common factor. The corresponding formula has the form NOC (a, b) \u003d ab: GCD (a, b) . Consider examples of finding NOCs using the above formula.

Example.

Find the smallest common multiple of two numbers 126 and 70.

Decision.

In this example, a \u003d 126, b \u003d 70. We use the connection of the NOC with the GCD, expressed by the formula NOC (a, b) \u003d ab: GCD (a, b). That is, first we have to find the greatest common divisor of the numbers 70 and 126, after which we can calculate the NOC of these numbers using the written formula.

We find the GCD (126, 70) using the Euclidean algorithm: 126 \u003d 701 + 56, 70 \u003d 561 + 14, 56 \u003d 14.4, therefore, GCD (126, 70) \u003d 14.

Now we find the required least common multiple: NOC (126, 70) \u003d 12670: GCD (126, 70) \u003d   12670: 14 \u003d 630.

Answer:

NOC (126, 70) \u003d 630.

Example.

What is the NOC equal to (68, 34)?

Decision.

Because 68 is divided entirely by 34, then GCD (68, 34) \u003d 34. Now we calculate the smallest common multiple: NOC (68, 34) \u003d 6834: GCD (68, 34) \u003d   6834: 34 \u003d 68.

Answer:

NOC (68, 34) \u003d 68.

Note that the previous example fits the following rule for finding LCL for positive integers a and b: if the number a is divisible by b, then the least common multiple of these numbers is a.

Finding NOCs by factoring numbers into prime factors

Another way to find the smallest common multiple is based on the decomposition of numbers into prime factors. If we compose a product of all prime factors of given numbers, and then exclude from this product all common prime factors that are present in the decompositions of these numbers, then the resulting product will be equal to the smallest common multiple of these numbers.

The stated rule for finding NOCs follows from the equality NOC (a, b) \u003d ab: GCD (a, b). Indeed, the product of numbers a and b is equal to the product of all factors involved in the decompositions of a and b. In its turn, GCD (a, b) is equal to the product of all prime factors simultaneously present in the decompositions of a and b (which is described in the section on finding GCD by decomposing numbers into prime factors).

We give an example. Let us know that 75 \u003d 3 · 5 · 5 and 210 \u003d 2 · 3 · 5 · 7. We compose the product of all factors of these expansions: 2 · 3 · 3 · 5 · 5 · 5 · 7. Now we exclude from this product all the factors that are present both in the expansion of the number 75 and in the expansion of the number 210 (3 and 5 are such factors), then the product will take the form 2 · 3 · 5 · 5 · 7. The value of this product is equal to the smallest common multiple of numbers 75 and 210, that is, NOC (75, 210) \u003d 2 · 3 · 5 · 5 · 7 \u003d 1,050.

Example.

Decomposing the numbers 441 and 700 into prime factors, find the smallest common multiple of these numbers.

Decision.

Factor 441 and 700 into prime factors:

We get 441 \u003d 3 · 3 · 7 · 7 and 700 \u003d 2 · 2 · 5 · 5 · 7.

Now we compose the product of all the factors involved in the decompositions of these numbers: 2 · 2 · 3 · 3 · 5 · 5 · 7 · 7 · 7. We exclude from this product all factors simultaneously present in both expansions (there is only one such factor - this is number 7): 2 · 2 · 3 · 3 · 5 · 5 · 7 · 7. Thus, NOC (441, 700) \u003d 2 · 2 · 3 · 3 · 5 · 5 · 7 · 7 \u003d 44 100.

Answer:

NOC (441, 700) \u003d 44 100.

The rule for finding NOCs using the decomposition of numbers into prime factors can be formulated a little differently. If we add the missing factors from the expansion of b to the factors from the expansion of the number a, then the value of the resulting product will be equal to the least common multiple of the numbers a and b.

For an example we will take all the same numbers 75 and 210, their decompositions into prime factors are as follows: 75 \u003d 3 · 5 · 5 and 210 \u003d 2 · 3 · 5 · 7. To the factors 3, 5 and 5 from the expansion of the number 75, we add the missing factors 2 and 7 from the expansion of the number 210, we obtain the product 2 · 3 · 5 · 5 · 7, the value of which is equal to the LCL (75, 210).

Example.

Find the smallest common multiple of 84 and 648.

Decision.

First we get the decompositions of the numbers 84 and 648 into prime factors. They have the form 84 \u003d 2 · 2 · 3 · 7 and 648 \u003d 2 · 2 · 2 · 3 · 3 · 3 · 3. To the factors 2, 2, 3, and 7 from the expansion of the number 84, we add the missing factors 2, 3, 3, and 3 from the expansion of the number 648, we obtain the product 2 · 2 · 2 · 3 · 3 · 3 · 3 · 7, which is 4 536 . Thus, the desired lowest common multiple of 84 and 648 is 4,536.

Answer:

NOC (84, 648) \u003d 4,536.

Finding NOCs of three or more numbers

The smallest common multiple of three or more numbers can be found by successively finding the NOC of two numbers. Recall the corresponding theorem, which gives a way to find the LCL of three or more numbers.

Theorem.

Let positive integers a 1, a 2, ..., ak be given, the smallest common multiple mk of these numbers can be found by sequentially calculating m 2 \u003d NOC (a 1, a 2), m 3 \u003d NOC (m 2, a 3), ... , mk \u003d LOC (mk − 1, ak).

Let us consider the application of this theorem by the example of finding the least common multiple of four numbers.

Example.

Find the NOC of the four numbers 140, 9, 54, and 250.

Decision.

In this example, a 1 \u003d 140, a 2 \u003d 9, a 3 \u003d 54, a 4 \u003d 250.

First we find m 2 \u003d NOC (a 1, a 2) \u003d NOC (140, 9). For this, according to the Euclidean algorithm, we determine the GCD (140, 9), we have 140 \u003d 9 · 15 + 5, 9 \u003d 5 · 1 + 4, 5 \u003d 4 · 1 + 1, 4 \u003d 1 · 4, therefore, the GCD (140, 9) \u003d 1, whence NOC (140, 9) \u003d 1409: NCD (140, 9) \u003d   1409: 1 \u003d 1,260. That is, m 2 \u003d 1,260.

Now find m 3 \u003d NOC (m 2, a 3) \u003d NOC (1 260, 54). We calculate it through the GCD (1 260, 54), which is also determined by the Euclidean algorithm: 1 260 \u003d 54 · 23 + 18, 54 \u003d 18 · 3. Then the GCD (1 260, 54) \u003d 18, whence the NOC (1,260, 54) \u003d 1,260 · 54: the GCD (1,260, 54) \u003d 1,260 · 54: 18 \u003d 3,780. That is, m 3 \u003d 3,780.

It remains to find m 4 \u003d NOC (m 3, a 4) \u003d NOC (3 780, 250). To do this, we find the GCD (3,780,250) according to the Euclidean algorithm: 3,780 \u003d 25015 + 30, 250 \u003d 30,8 + 10, 30 \u003d 10,3. Therefore, GCD (3,780,250) \u003d 10, whence NOC (3,780,250) \u003d 3,780250: GCD (3,780,250) \u003d   3,780,250: 10 \u003d 94,500. That is, m 4 \u003d 94 500.

Thus, the smallest common multiple of the original four numbers is 94,500.

Answer:

NOC (140, 9, 54, 250) \u003d 94 500.

In many cases, the least common multiple of three or more numbers is conveniently found using factorizations of these numbers. The following rule should be followed. The smallest common multiple of several numbers is equal to the product, which is composed as follows: to all factors from the expansion of the first number, the missing factors from the expansion of the second number are added, to the resulting factors are added the missing factors from the expansion of the third number, and so on.

Consider the example of finding the smallest common multiple using a prime factorization of numbers.

Example.

Find the smallest common multiple of five numbers 84, 6, 48, 7, 143.

Decision.

First we get the decompositions of these numbers into prime factors: 84 \u003d 2 · 2 · 3 · 7, 6 \u003d 2 · 3, 48 \u003d 2 · 2 · 2 · 2 · 3, 7 (7 is a prime number, it coincides with its factorization prime factors) and 143 \u003d 11 · 13.

To find the NOC of these numbers to the factors of the first number 84 (they are 2, 2, 3 and 7), you need to add the missing factors from the expansion of the second number 6. The expansion of 6 does not contain the missing factors, since 2 and 3 are already present in the expansion of the first number 84. Further to the factors 2, 2, 3 and 7, we add the missing factors 2 and 2 from the expansion of the third number 48, we get a set of factors 2, 2, 2, 2, 3, and 7. You will not have to add factors to this set in the next step, since 7 is already contained in it. Finally, to the factors 2, 2, 2, 2, 3, and 7, we add the missing factors 11 and 13 from the expansion of 143. We get the product 2 · 2 · 2 · 2 · 3 · 7 · 11 · 13, which is 48 048.