Finding the least common multiple: ways, examples of finding the LCM. Common divisor and multiple

Greatest Common Divisor

Definition 2

If a natural number a is divisible by a natural number $b$, then $b$ is called a divisor of $a$, and the number $a$ is called a multiple of $b$.

Let $a$ and $b$ be natural numbers. The number $c$ is called a common divisor for both $a$ and $b$.

The set of common divisors of the numbers $a$ and $b$ is finite, since none of these divisors can be greater than $a$. This means that among these divisors there is the largest one, which is called the greatest common divisor of the numbers $a$ and $b$, and the notation is used to denote it:

$gcd \ (a;b) \ ​​or \ D \ (a;b)$

To find the greatest common divisor of two numbers:

1. Find the product of the numbers found in step 2. The resulting number will be the desired greatest common divisor.

Example 1

Find the gcd of the numbers $121$ and $132.$

$242=2\cdot 11\cdot 11$

$132=2\cdot 2\cdot 3\cdot 11$

Choose the numbers that are included in the expansion of these numbers

$242=2\cdot 11\cdot 11$

$132=2\cdot 2\cdot 3\cdot 11$

Find the product of the numbers found in step 2. The resulting number will be the desired greatest common divisor.

$gcd=2\cdot 11=22$

Example 2

Find the GCD of monomials $63$ and $81$.

We will find according to the presented algorithm. For this:

Let's decompose numbers into prime factors

$63=3\cdot 3\cdot 7$

$81=3\cdot 3\cdot 3\cdot 3$

We select the numbers that are included in the expansion of these numbers

$63=3\cdot 3\cdot 7$

$81=3\cdot 3\cdot 3\cdot 3$

Let's find the product of the numbers found in step 2. The resulting number will be the desired greatest common divisor.

$gcd=3\cdot 3=9$

You can find the GCD of two numbers in another way, using the set of divisors of numbers.

Example 3

Find the gcd of the numbers $48$ and $60$.

Solution:

Find the set of divisors of $48$: $\left\((\rm 1,2,3.4.6,8,12,16,24,48)\right\)$

Now let's find the set of divisors of $60$:$\ \left\((\rm 1,2,3,4,5,6,10,12,15,20,30,60)\right\)$

Let's find the intersection of these sets: $\left\((\rm 1,2,3,4,6,12)\right\)$ - this set will determine the set of common divisors of the numbers $48$ and $60$. The largest element in this set will be the number $12$. So the greatest common divisor of $48$ and $60$ is $12$.

Definition of NOC

Definition 3

common multiple of natural numbers$a$ and $b$ is a natural number that is a multiple of both $a$ and $b$.

Common multiples of numbers are numbers that are divisible by the original without a remainder. For example, for the numbers $25$ and $50$, the common multiples will be the numbers $50,100,150,200$, etc.

The least common multiple will be called the least common multiple and denoted by LCM$(a;b)$ or K$(a;b).$

To find the LCM of two numbers, you need:

1. Decompose numbers into prime factors
2. Write out the factors that are part of the first number and add to them the factors that are part of the second and do not go to the first

Example 4

Find the LCM of the numbers $99$ and $77$.

We will find according to the presented algorithm. For this

Decompose numbers into prime factors

$99=3\cdot 3\cdot 11$

Write down the factors included in the first

add to them factors that are part of the second and do not go to the first

Find the product of the numbers found in step 2. The resulting number will be the desired least common multiple

$LCC=3\cdot 3\cdot 11\cdot 7=693$

Compiling lists of divisors of numbers is often very time consuming. There is a way to find GCD called Euclid's algorithm.

Statements on which Euclid's algorithm is based:

If $a$ and $b$ are natural numbers, and $a\vdots b$, then $D(a;b)=b$

If $a$ and $b$ are natural numbers such that $b

Using $D(a;b)= D(a-b;b)$, we can successively decrease the numbers under consideration until we reach a pair of numbers such that one of them is divisible by the other. Then the smaller of these numbers will be the desired greatest common divisor for the numbers $a$ and $b$.

Properties of GCD and LCM

1. Any common multiple of $a$ and $b$ is divisible by K$(a;b)$
2. If $a\vdots b$ , then K$(a;b)=a$

If K$(a;b)=k$ and $m$-natural number, then K$(am;bm)=km$

If $d$ is a common divisor for $a$ and $b$, then K($\frac(a)(d);\frac(b)(d)$)=$\ \frac(k)(d) $

If $a\vdots c$ and $b\vdots c$ , then $\frac(ab)(c)$ is a common multiple of $a$ and $b$

For any natural numbers $a$ and $b$ the equality

$D(a;b)\cdot K(a;b)=ab$

Any common divisor of $a$ and $b$ is a divisor of $D(a;b)$

Let's continue the discussion about the least common multiple that we started in the LCM - Least Common Multiple, Definition, Examples section. In this topic, we will look at ways to find the LCM for three numbers or more, we will analyze the question of how to find the LCM of a negative number.

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Calculation of the least common multiple (LCM) through gcd

We have already established the relationship between the least common multiple and the greatest common divisor. Now let's learn how to define the LCM through the GCD. First, let's figure out how to do this for positive numbers.

Definition 1

You can find the least common multiple through the greatest common divisor using the formula LCM (a, b) \u003d a b: GCD (a, b) .

Example 1

It is necessary to find the LCM of the numbers 126 and 70.

Solution

Let's take a = 126 , b = 70 . Substitute the values ​​in the formula for calculating the least common multiple through the greatest common divisor LCM (a, b) = a · b: GCD (a, b) .

Finds the GCD of the numbers 70 and 126. For this we need the Euclid algorithm: 126 = 70 1 + 56 , 70 = 56 1 + 14 , 56 = 14 4 , hence gcd (126 , 70) = 14 .

Let's calculate the LCM: LCM (126, 70) = 126 70: GCD (126, 70) = 126 70: 14 = 630.

Answer: LCM (126, 70) = 630.

Example 2

Find the nok of the numbers 68 and 34.

Solution

GCD in this case is easy to find, since 68 is divisible by 34. Calculate the least common multiple using the formula: LCM (68, 34) = 68 34: GCD (68, 34) = 68 34: 34 = 68.

Answer: LCM(68, 34) = 68.

In this example, we used the rule for finding the least common multiple of 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 the LCM by Factoring Numbers into Prime Factors

Now let's look at a way to find the LCM, 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 make up the product of all prime factors of numbers for which we need to find the LCM;
• we exclude all prime factors from their obtained products;
• the product obtained after eliminating the common prime factors will be equal to the LCM of the given numbers.

This way of finding the least common multiple is based on the equality LCM (a , b) = a b: GCD (a , b) . If you look at the formula, it will become clear: the product of the numbers a and b is equal to the product of all factors that are involved in the expansion of these two numbers. In this case, 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 out like this: 75 = 3 5 5 and 210 = 2 3 5 7. If you make the product of all the factors of the two original numbers, you get: 2 3 3 5 5 5 7.

If we exclude the factors common to both numbers 3 and 5, we get a product of the following form: 2 3 5 5 7 = 1050. This product will be our LCM for the numbers 75 and 210.

Example 4

Find the LCM of numbers 441 and 700 , decomposing both numbers into prime factors.

Solution

Let's 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 = 3 3 7 7 and 700 = 2 2 5 5 7 .

The product of all the factors that participated in the expansion of these numbers will look like: 2 2 3 3 5 5 7 7 7. Let's find the common factors. This number is 7 . We exclude it from the general product: 2 2 3 3 5 5 7 7. It turns out that NOC (441 , 700) = 2 2 3 3 5 5 7 7 = 44 100.

Answer: LCM (441 , 700) = 44 100 .

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

Definition 3

Previously, we excluded from the total number of factors common to both numbers. Now we will do it differently:

• Let's decompose both numbers into prime factors:
• add to the product of the prime factors of the first number the missing factors of the second number;
• we get the product, which will be the desired LCM of two numbers.

Example 5

Let's go back to the numbers 75 and 210 , for which we already looked for the LCM in one of the previous examples. Let's break them down into simple factors: 75 = 3 5 5 and 210 = 2 3 5 7. To the product of factors 3 , 5 and 5 number 75 add the missing factors 2 and 7 numbers 210 . We get: 2 3 5 5 7 . This is the LCM of the numbers 75 and 210.

Example 6

It is necessary to calculate the LCM of the numbers 84 and 648.

Solution

Let's decompose the numbers from the condition into prime factors: 84 = 2 2 3 7 and 648 = 2 2 2 3 3 3 3. Add to the product of the factors 2 , 2 , 3 and 7 numbers 84 missing factors 2 , 3 , 3 and
3 numbers 648 . We get the product 2 2 2 3 3 3 3 7 = 4536 . This is the least common multiple of 84 and 648.

Answer: LCM (84, 648) = 4536.

Finding the LCM 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 consistently find the LCM 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 of these numbers is found in sequential calculation m 2 = LCM (a 1 , a 2) , m 3 = LCM (m 2 , a 3) , … , m k = LCM (m k − 1 , a k) .

Now let's look at how the theorem can be applied to specific problems.

Example 7

You need to calculate the least common multiple of the four numbers 140 , 9 , 54 and 250 .

Solution

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

Let's start by calculating m 2 = LCM (a 1 , a 2) = LCM (140 , 9) . Let's use the Euclidean algorithm to calculate the GCD of the numbers 140 and 9: 140 = 9 15 + 5 , 9 = 5 1 + 4 , 5 = 4 1 + 1 , 4 = 1 4 . We get: GCD(140, 9) = 1, LCM(140, 9) = 140 9: GCD(140, 9) = 140 9: 1 = 1260. Therefore, m 2 = 1 260 .

Now let's calculate according to the same algorithm m 3 = LCM (m 2 , a 3) = LCM (1 260 , 54) . In the course of calculations, we get m 3 = 3 780.

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

The LCM of the four numbers from the example condition is 94500 .

Answer: LCM (140, 9, 54, 250) = 94,500.

As you can see, the calculations are simple, but quite laborious. To save time, you can go the other way.

Definition 4

We offer you the following algorithm of actions:

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

Example 8

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

Solution

Let's decompose all five numbers into prime factors: 84 = 2 2 3 7 , 6 = 2 3 , 48 = 2 2 2 2 3 , 7 , 143 = 11 13 . prime numbers, which is the number 7 , cannot be factored into prime factors. Such numbers coincide with their decomposition into prime factors.

Now let's take the product of the prime factors 2, 2, 3 and 7 of the number 84 and add to them the missing factors of the second number. We have decomposed 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 multipliers. We turn to the number 48, from the product of prime factors of which we take 2 and 2. Then we add a simple factor of 7 from the fourth number and factors of 11 and 13 of the fifth. We get: 2 2 2 2 3 7 11 13 = 48,048. This is the least common multiple of the five original numbers.

Answer: LCM (84, 6, 48, 7, 143) = 48,048.

Finding the Least Common Multiple of Negative Numbers

To find the least common multiple negative numbers, these numbers must first be replaced by numbers with the opposite sign, and then the calculations should be carried out according to the above algorithms.

Example 9

LCM(54, −34) = LCM(54, 34) and LCM(−622,−46, −54,−888) = LCM(622, 46, 54, 888) .

Such actions are permissible due to the fact that if it is accepted that a and − a- opposite numbers
then the set of multiples a coincides with the set of multiples of a number − a.

Example 10

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

Solution

Let's change the numbers − 145 and − 45 to their opposite numbers 145 and 45 . Now, using the algorithm, we calculate the LCM (145 , 45) = 145 45: GCD (145 , 45) = 145 45: 5 = 1 305 , having previously determined the GCD using the Euclid algorithm.

We get that the LCM of numbers − 145 and − 45 equals 1 305 .

Answer: LCM (− 145 , − 45) = 1 305 .

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Second number: b=

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Result:

Greatest Common Divisor gcd( a,b)=6

Least common multiple of LCM( a,b)=468

The largest natural number by which the numbers a and b are divisible without remainder is called greatest common divisor(gcd) of these numbers. Denoted gcd(a,b), (a,b), gcd(a,b) or hcf(a,b).

Least common multiple(LCM) of two integers a and b is the smallest natural number that is divisible by a and b without a remainder. Denoted LCM(a,b), or lcm(a,b).

Integers a and b are called coprime if they have no common divisors other than +1 and −1.

Greatest Common Divisor

Let two positive numbers be given a 1 and a 2 1). It is required to find a common divisor of these numbers, i.e. find such a number λ , which divides the numbers a 1 and a 2 at the same time. Let's describe the algorithm.

1) In this article, the word number will mean an integer.

Let a 1 ≥ a 2 and let

where m 1 , a 3 are some integers, a 3 <a 2 (remainder from division a 1 on a 2 should be less a 2).

Let's pretend that λ divides a 1 and a 2 , then λ divides m 1 a 2 and λ divides a 1 −m 1 a 2 =a 3 (Assertion 2 of the article "Divisibility of numbers. Sign of divisibility"). It follows that every common divisor a 1 and a 2 is a common divisor a 2 and a 3 . The converse is also true if λ common divisor a 2 and a 3 , then m 1 a 2 and a 1 =m 1 a 2 +a 3 are also divided into λ . Hence the common divisor a 2 and a 3 is also a common divisor a 1 and a 2. Because a 3 <a 2 ≤a 1 , then we can say that the solution to the problem of finding a common divisor of numbers a 1 and a 2 reduced to a simpler problem of finding a common divisor of numbers a 2 and a 3 .

If a a 3 ≠0, then we can divide a 2 on a 3 . Then

,

where m 1 and a 4 are some integers, ( a 4 remainder of division a 2 on a 3 (a 4 <a 3)). By similar reasoning, we come to the conclusion that the common divisors of numbers a 3 and a 4 is the same as common divisors of numbers a 2 and a 3 , and also with common divisors a 1 and a 2. Because a 1 , a 2 , a 3 , a 4 , ... numbers that are constantly decreasing, and since there is a finite number of integers between a 2 and 0, then at some step n, remainder of the division a n on a n+1 will be equal to zero ( a n+2=0).

.

Every common divisor λ numbers a 1 and a 2 is also a divisor of numbers a 2 and a 3 , a 3 and a 4 , .... a n and a n+1 . The converse is also true, common divisors of numbers a n and a n+1 are also divisors of numbers a n−1 and a n , .... , a 2 and a 3 , a 1 and a 2. But the common divisor a n and a n+1 is a number a n+1 , because a n and a n+1 are divisible by a n+1 (recall that a n+2=0). Consequently a n+1 is also a divisor of numbers a 1 and a 2 .

Note that the number a n+1 is the greatest number divisor a n and a n+1 , since the greatest divisor a n+1 is itself a n+1 . If a a n + 1 can be represented as a product of integers, then these numbers are also common divisors of numbers a 1 and a 2. Number a n+1 are called greatest common divisor numbers a 1 and a 2 .

Numbers a 1 and a 2 can be both positive and negative numbers. If one of the numbers is equal to zero, then the greatest common divisor of these numbers will be equal to the absolute value of the other number. The greatest common divisor of zero numbers is not defined.

The above algorithm is called Euclid's algorithm to find the greatest common divisor of two integers.

An example of finding the greatest common divisor of two numbers

Find the greatest common divisor of two numbers 630 and 434.

• Step 1. Divide the number 630 by 434. The remainder is 196.
• Step 2. Divide the number 434 by 196. The remainder is 42.
• Step 3. Divide the number 196 by 42. The remainder is 28.
• Step 4. Divide the number 42 by 28. The remainder is 14.
• Step 5. Divide the number 28 by 14. The remainder is 0.

At step 5, the remainder of the division is 0. Therefore, the greatest common divisor of the numbers 630 and 434 is 14. Note that the numbers 2 and 7 are also divisors of the numbers 630 and 434.

Coprime numbers

Definition 1. Let the greatest common divisor of numbers a 1 and a 2 is equal to one. Then these numbers are called coprime numbers that do not have a common divisor.

Theorem 1. If a a 1 and a 2 relatively prime numbers, and λ some number, then any common divisor of numbers λa 1 and a 2 is also a common divisor of numbers λ and a 2 .

Proof. Consider Euclid's algorithm for finding the greatest common divisor of numbers a 1 and a 2 (see above).

.

It follows from the conditions of the theorem that the greatest common divisor of numbers a 1 and a 2 , and therefore a n and a n+1 is 1. I.e. a n+1=1.

Let's multiply all these equalities by λ , then

.

Let the common divisor a 1 λ and a 2 is δ . Then δ enters as a factor in a 1 λ , m 1 a 2 λ and in a 1 λ -m 1 a 2 λ =a 3 λ (See "Divisibility of numbers", Statement 2). Further δ enters as a factor in a 2 λ and m 2 a 3 λ , and hence enters as a factor in a 2 λ -m 2 a 3 λ =a 4 λ .

By reasoning in this way, we are convinced that δ enters as a factor in a n−1 λ and m n−1 a n λ , and therefore in a n−1 λ −m n−1 a n λ =a n+1 λ . Because a n+1 =1, then δ enters as a factor in λ . Hence the number δ is a common divisor of numbers λ and a 2 .

Consider special cases of Theorem 1.

Consequence 1. Let a and c prime numbers are relatively b. Then their product ac is a prime number with respect to b.

Really. From Theorem 1 ac and b have the same common divisors as c and b. But the numbers c and b coprime, i.e. have a single common divisor 1. Then ac and b also have a single common divisor 1. Hence ac and b mutually simple.

Consequence 2. Let a and b coprime numbers and let b divides ak. Then b divides and k.

Really. From the assertion condition ak and b have a common divisor b. By virtue of Theorem 1, b must be a common divisor b and k. Consequently b divides k.

Corollary 1 can be generalized.

Consequence 3. 1. Let the numbers a 1 , a 2 , a 3 , ..., a m are prime relative to the number b. Then a 1 a 2 , a 1 a 2 · a 3 , ..., a 1 a 2 a 3 ··· a m , the product of these numbers is prime with respect to the number b.

2. Let we have two rows of numbers

such that every number in the first row is prime with respect to every number in the second row. Then the product

It is required to find such numbers that are divisible by each of these numbers.

If the number is divisible by a 1 , then it looks like sa 1 , where s some number. If a q is the greatest common divisor of numbers a 1 and a 2 , then

where s 1 is some integer. Then

is least common multiple of numbers a 1 and a 2 .

a 1 and a 2 coprime, then the least common multiple of the numbers a 1 and a 2:

Find the least common multiple of these numbers.

It follows from the above that any multiple of the numbers a 1 , a 2 , a 3 must be a multiple of numbers ε and a 3 and vice versa. Let the least common multiple of the numbers ε and a 3 is ε one . Further, a multiple of numbers a 1 , a 2 , a 3 , a 4 must be a multiple of numbers ε 1 and a four . Let the least common multiple of the numbers ε 1 and a 4 is ε 2. Thus, we found out that all multiples of numbers a 1 , a 2 , a 3 ,...,a m coincide with multiples of some specific number ε n , which is called the least common multiple of the given numbers.

In the particular case when the numbers a 1 , a 2 , a 3 ,...,a m coprime, then the least common multiple of the numbers a 1 , a 2 as shown above has the form (3). Further, since a 3 prime with respect to numbers a 1 , a 2 , then a 3 is a prime relative number a one · a 2 (Corollary 1). So the least common multiple of the numbers a 1 ,a 2 ,a 3 is a number a one · a 2 · a 3 . Arguing in a similar way, we arrive at the following assertions.

Statement 1. Least common multiple of coprime numbers a 1 , a 2 , a 3 ,...,a m is equal to their product a one · a 2 · a 3 ··· a m .

Statement 2. Any number that is divisible by each of the coprime numbers a 1 , a 2 , a 3 ,...,a m is also divisible by their product a one · a 2 · a 3 ··· a m .

A multiple of a number is a number that is divisible by a given number without a remainder. The least common multiple (LCM) of a group of numbers is the smallest number that is evenly divisible by each number in the group. To find the least common multiple, you need to find the prime factors of the given numbers. Also, LCM can be calculated using a number of other methods that are applicable to groups of two or more numbers.

Steps

A series of multiples

Look at these numbers. The method described here is best used when given two numbers that are both less than 10. If large numbers are given, use a different method.

• For example, find the least common multiple of the numbers 5 and 8. These are small numbers, so this method can be used.

1. A multiple of a number is a number that is divisible by a given number without a remainder. Multiple numbers can be found in the multiplication table.

   • For example, numbers that are multiples of 5 are: 5, 10, 15, 20, 25, 30, 35, 40.

2. Write down a series of numbers that are multiples of the first number. Do this under multiples of the first number to compare two rows of numbers.

   • For example, numbers that are multiples of 8 are: 8, 16, 24, 32, 40, 48, 56, and 64.

3. Find the smallest number that appears in both series of multiples. You may have to write long series of multiples to find the total. The smallest number that appears in both series of multiples is the least common multiple.

   • For example, the smallest number that appears in the series of multiples of 5 and 8 is 40. Therefore, 40 is the least common multiple of 5 and 8.

   Prime factorization

   1. Look at these numbers. The method described here is best used when given two numbers that are both greater than 10. If smaller numbers are given, use a different method.

      • For example, find the least common multiple of the numbers 20 and 84. Each of the numbers is greater than 10, so this method can be used.

   2. Factorize the first number. That is, you need to find such prime numbers, when multiplied, you get a given number. Having found prime factors, write them down as an equality.

      • For example, 2 × 10 = 20 (\displaystyle (\mathbf (2) )\times 10=20) and 2 × 5 = 10 (\displaystyle (\mathbf (2) )\times (\mathbf (5) )=10). Thus, the prime factors of the number 20 are the numbers 2, 2 and 5. Write them down as an expression: .

   3. Factor the second number into prime factors. Do this in the same way as you factored the first number, that is, find such prime numbers that, when multiplied, will get this number.

      • For example, 2 × 42 = 84 (\displaystyle (\mathbf (2) )\times 42=84), 7 × 6 = 42 (\displaystyle (\mathbf (7) )\times 6=42) and 3 × 2 = 6 (\displaystyle (\mathbf (3) )\times (\mathbf (2) )=6). Thus, the prime factors of the number 84 are the numbers 2, 7, 3 and 2. Write them down as an expression: .

   4. Write down the factors common to both numbers. Write such factors as a multiplication operation. As you write down each factor, cross it out in both expressions (expressions that describe the decomposition of numbers into prime factors).

      • For example, the common factor for both numbers is 2, so write 2 × (\displaystyle 2\times ) and cross out the 2 in both expressions.
      • The common factor for both numbers is another factor of 2, so write 2 × 2 (\displaystyle 2\times 2) and cross out the second 2 in both expressions.

   5. Add the remaining factors to the multiplication operation. These are factors that are not crossed out in both expressions, that is, factors that are not common to both numbers.

      • For example, in the expression 20 = 2 × 2 × 5 (\displaystyle 20=2\times 2\times 5) both twos (2) are crossed out because they are common factors. The factor 5 is not crossed out, so write the multiplication operation as follows: 2 × 2 × 5 (\displaystyle 2\times 2\times 5)
      • In the expression 84 = 2 × 7 × 3 × 2 (\displaystyle 84=2\times 7\times 3\times 2) both deuces (2) are also crossed out. Factors 7 and 3 are not crossed out, so write the multiplication operation as follows: 2 × 2 × 5 × 7 × 3 (\displaystyle 2\times 2\times 5\times 7\times 3).

   6. Calculate the least common multiple. To do this, multiply the numbers in the written multiplication operation.

      • For example, 2 × 2 × 5 × 7 × 3 = 420 (\displaystyle 2\times 2\times 5\times 7\times 3=420). So the least common multiple of 20 and 84 is 420.

   Finding common divisors

   1. Draw a grid like you would for a game of tic-tac-toe. Such a grid consists of two parallel lines that intersect (at right angles) with two other parallel lines. This will result in three rows and three columns (the grid looks a lot like the # sign). Write the first number in the first row and second column. Write the second number in the first row and third column.

      • For example, find the least common multiple of 18 and 30. Write 18 in the first row and second column, and write 30 in the first row and third column.

   2. Find the divisor common to both numbers. Write it down in the first row and first column. It is better to look for prime divisors, but this is not a prerequisite.

      • For example, 18 and 30 are even numbers, so their common divisor is 2. So write 2 in the first row and first column.

   3. Divide each number by the first divisor. Write each quotient under the corresponding number. The quotient is the result of dividing two numbers.

      • For example, 18 ÷ 2 = 9 (\displaystyle 18\div 2=9), so write 9 under 18.
      • 30 ÷ 2 = 15 (\displaystyle 30\div 2=15), so write 15 under 30.

   4. Find a divisor common to both quotients. If there is no such divisor, skip the next two steps. Otherwise, write down the divisor in the second row and first column.

      • For example, 9 and 15 are divisible by 3, so write 3 in the second row and first column.

   5. Divide each quotient by the second divisor. Write each division result under the corresponding quotient.

      • For example, 9 ÷ 3 = 3 (\displaystyle 9\div 3=3), so write 3 under 9.
      • 15 ÷ 3 = 5 (\displaystyle 15\div 3=5), so write 5 under 15.

   6. If necessary, supplement the grid with additional cells. Repeat the above steps until the quotients have a common divisor.

   7. Circle the numbers in the first column and last row of the grid. Then write the highlighted numbers as a multiplication operation.

      • For example, the numbers 2 and 3 are in the first column, and the numbers 3 and 5 are in the last row, so write the multiplication operation like this: 2 × 3 × 3 × 5 (\displaystyle 2\times 3\times 3\times 5).

   8. Find the result of multiplying numbers. This will calculate the least common multiple of the two given numbers.

      • For example, 2 × 3 × 3 × 5 = 90 (\displaystyle 2\times 3\times 3\times 5=90). So the least common multiple of 18 and 30 is 90.

   Euclid's algorithm

   1. Remember the terminology associated with the division operation. The dividend is the number that is being divided. The divisor is the number by which to divide. The quotient is the result of dividing two numbers. The remainder is the number left when two numbers are divided.

      • For example, in the expression 15 ÷ 6 = 2 (\displaystyle 15\div 6=2) rest. 3:
        15 is the divisible
        6 is the divisor
        2 is private
        3 is the remainder.

Students are given a lot of math assignments. Among them, very often there are tasks with the following formulation: there are two values. How to find the least common multiple of 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 LCM and the basic concepts.

Before finding the answer to the question of how to find the LCM, you need to define the term multiple. Most often, the formulation of this concept is as follows: a multiple of some value A is a natural number that will be divisible by A without a remainder. So, for 4, 8, 12, 16, 20 and so on, up to the required limit.

In this case, the number of divisors for a particular value can be limited, and there are infinitely many multiples. There is also the same value for natural values. This is an indicator that is divided by them without a remainder. Having dealt with the concept of the smallest value for certain indicators, let's move on to how to find it.

Finding the NOC

The least multiple of two or more exponents is the smallest natural number that is fully divisible by all the given numbers.

There are several ways to find such a value. Let's consider the following methods:

1. If the numbers are small, then write in the line all divisible by it. Keep doing this until you find something in 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 these are large or you need to find a multiple for 3 or more values, then here you should use a different technique that involves decomposing numbers into prime factors. First, lay out the largest of the indicated, then all the rest. Each of them has its own number of multipliers. As an example, let's decompose 20 (2*2*5) and 50 (5*5*2). For the smaller of them, underline the factors and add to the largest. The result will be 100, which will be the least common multiple of the above numbers.
3. When finding 3 numbers (16, 24 and 36) the principles are the same as for the other two. Let's expand each of them: 16 = 2*2*2*2, 24=2*2*2*3, 36=2*2*3*3. Only two deuces from the decomposition of the number 16 were not included in the expansion of the largest. We add them and get 144, which is the smallest result for the previously indicated numerical values.

Now we know what is the general technique for finding the smallest value for two, three or more values. However, there are also private methods, helping to search for NOCs, if the previous ones do not help.

How to find GCD and NOC.

Private Ways of Finding

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

• if one of the numbers is divisible by the others without a remainder, then the lowest multiple of these numbers is equal to it (NOC 60 and 15 is equal to 15);
• Coprime numbers do not have common prime divisors. Their smallest value is equal to the product of these numbers. Thus, for the numbers 7 and 8, this will be 56;
• the same rule works for other cases, including special ones, which can be read about in specialized literature. This should also include cases of decomposition of composite numbers, which are the subject of separate articles and even Ph.D. dissertations.

Special cases are less common than standard examples. But thanks to them, you can learn how 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 you can understand the principle of finding the smallest multiple:

1. We find LCM (35; 40). We lay out first 35 = 5*7, then 40 = 5*8. We add 8 to the smallest number and get the NOC 280.
2. NOC (45; 54). We lay out each of them: 45 = 3*3*5 and 54 = 3*3*6. We 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.

Finding the NOC is much easier than it might seem at first. For this, both a simple expansion and the multiplication of simple values ​​\u200b\u200bto 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 with different methods, this develops the logical apparatus and allows you to remember numerous terms. Learn methods for finding such an indicator and you will be able to work well with the rest of the mathematical sections. Happy learning math!

Video

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