The material presented below is a logical continuation of the theory from the article entitled LCM - least common multiple, definition, examples, connection between LCM and GCD. Here we will talk about finding the least common multiple (LCM), And Special attention Let's focus on solving examples. First, we will show how the LCM of two numbers is calculated using the GCD of these numbers. Next, we'll look at finding the least common multiple by factoring numbers into prime factors. After this, we will focus on finding the LCM of three or more numbers, and also pay attention to calculating the LCM of negative numbers.
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Calculating Least Common Multiple (LCM) via GCD
One way to find the least common multiple is based on the relationship between LCM and GCD. Existing connection between LCM and GCD allows you to calculate the least common multiple of two positive integers using a known greatest common divisor. The corresponding formula is LCM(a, b)=a b:GCD(a, b) . Let's look at examples of finding the LCM using the given formula.
Example.
Find the least common multiple of two numbers 126 and 70.
Solution.
In this example a=126 , b=70 . Let us use the connection between LCM and GCD, expressed by the formula LCM(a, b)=a b: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 LCM of these numbers using the written formula.
Let's find GCD(126, 70) using the Euclidean algorithm: 126=70·1+56, 70=56·1+14, 56=14·4, therefore, GCD(126, 70)=14.
Now we find the required least common multiple: GCD(126, 70)=126·70:GCD(126, 70)= 126·70:14=630.
Answer:
LCM(126, 70)=630 .
Example.
What is LCM(68, 34) equal to?
Solution.
Because 68 is divisible by 34, then GCD(68, 34)=34. Now we calculate the least common multiple: GCD(68, 34)=68·34:GCD(68, 34)= 68·34:34=68.
Answer:
LCM(68, 34)=68 .
Note that the previous example fits the following rule for finding the LCM 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 the LCM by factoring numbers into prime factors
Another way to find the least common multiple is based on factoring numbers into prime factors. If you compose a product from all the prime factors of given numbers, and then exclude from this product all the common prime factors present in the decompositions of the given numbers, then the resulting product will be equal to the least common multiple of the given numbers.
The stated rule for finding the LCM follows from the equality LCM(a, b)=a b:GCD(a, b). Indeed, the product of numbers a and b is equal to the product of all factors involved in the expansion of numbers a and b. In turn, GCD(a, b) is equal to the product of all prime factors simultaneously present in the expansions of numbers a and b (as described in the section on finding GCD using the expansion of numbers into prime factors).
Let's give an example. Let us know that 75=3·5·5 and 210=2·3·5·7. Let's compose the product from all the factors of these expansions: 2·3·3·5·5·5·7 . Now from this product we exclude all the factors present in both the expansion of the number 75 and the expansion of the number 210 (these factors are 3 and 5), then the product will take the form 2·3·5·5·7. The value of this product is equal to the least common multiple of 75 and 210, that is, NOC(75, 210)= 2·3·5·5·7=1,050.
Example.
Factor the numbers 441 and 700 into prime factors and find the least common multiple of these numbers.
Solution.
Let's factor the numbers 441 and 700 into prime factors:
We get 441=3·3·7·7 and 700=2·2·5·5·7.
Now let’s create a product from all the factors involved in the expansion of these numbers: 2·2·3·3·5·5·7·7·7. Let us exclude from this product all factors that are simultaneously present in both expansions (there is only one such factor - this is the number 7): 2·2·3·3·5·5·7·7. Thus, LCM(441, 700)=2·2·3·3·5·5·7·7=44 100.
Answer:
NOC(441, 700)= 44 100 .
The rule for finding the LCM using factorization of numbers into prime factors can be formulated a little differently. If the missing factors from the expansion of number b are added 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 example, let's take the same numbers 75 and 210, their decompositions into prime factors are as follows: 75=3·5·5 and 210=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 LCM(75, 210).
Example.
Find the least common multiple of 84 and 648.
Solution.
We first obtain the decompositions of the numbers 84 and 648 into prime factors. They look like 84=2·2·3·7 and 648=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 equal to 4 536 . Thus, the desired least common multiple of 84 and 648 is 4,536.
Answer:
LCM(84, 648)=4,536 .
Finding the LCM of three or more numbers
The least common multiple of three or more numbers can be found by sequentially finding the LCM of two numbers. Let us recall the corresponding theorem, which gives a way to find the LCM of three or more numbers.
Theorem.
Let positive integer numbers a 1 , a 2 , …, a k be given, the least common multiple m k of these numbers is found by sequentially calculating m 2 = LCM(a 1 , a 2) , m 3 = LCM(m 2 , a 3) , … , m k = LCM(m k−1 , a k) .
Let's consider the application of this theorem using the example of finding the least common multiple of four numbers.
Example.
Find the LCM of four numbers 140, 9, 54 and 250.
Solution.
In this example, a 1 =140, a 2 =9, a 3 =54, a 4 =250.
First we find m 2 = LOC(a 1 , a 2) = LOC(140, 9). To do this, using the Euclidean algorithm, we determine GCD(140, 9), we have 140=9·15+5, 9=5·1+4, 5=4·1+1, 4=1·4, therefore, GCD(140, 9)=1 , from where GCD(140, 9)=140 9:GCD(140, 9)= 140·9:1=1,260. That is, m 2 =1 260.
Now we find m 3 = LOC (m 2 , a 3) = LOC (1 260, 54). Let's calculate it through GCD(1 260, 54), which we also determine using the Euclidean algorithm: 1 260=54·23+18, 54=18·3. Then gcd(1,260, 54)=18, from which gcd(1,260, 54)= 1,260·54:gcd(1,260, 54)= 1,260·54:18=3,780. That is, m 3 =3 780.
All that remains is to find m 4 = LOC(m 3, a 4) = LOC(3 780, 250). To do this, we find GCD(3,780, 250) using the Euclidean algorithm: 3,780=250·15+30, 250=30·8+10, 30=10·3. Therefore, GCM(3,780, 250)=10, whence GCM(3,780, 250)= 3 780 250: GCD(3 780, 250)= 3,780·250:10=94,500. That is, m 4 =94,500.
So the least common multiple of the original four numbers is 94,500.
Answer:
LCM(140, 9, 54, 250)=94,500.
In many cases, it is convenient to find the least common multiple of three or more numbers using prime factorizations of the given numbers. In this case, you should adhere to next rule. The least common multiple of several numbers is equal to the product, which is composed as follows: the missing factors from the expansion of the second number are added to all the factors from the expansion of the first number, the missing factors from the expansion of the third number are added to the resulting factors, and so on.
Let's look at an example of finding the least common multiple using prime factorization.
Example.
Find the least common multiple of the five numbers 84, 6, 48, 7, 143.
Solution.
First, we obtain decompositions of these numbers into prime factors: 84=2·2·3·7, 6=2·3, 48=2·2·2·2·3, 7 (7 is a prime number, it coincides with its decomposition into prime factors) and 143=11·13.
To find the LCM 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 decomposition of the number 6 does not contain missing factors, since both 2 and 3 are already present in the decomposition of the first number 84. Next, 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. There will be no need to add multipliers 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 the number 143. We get the product 2·2·2·2·3·7·11·13, which is equal to 48,048.
Online calculator allows you to quickly find the greatest common divisor and least common multiple of both two and any other number of numbers.
Calculator for finding GCD and LCM
Find GCD and LOC
Found GCD and LOC: 5806
How to use the calculator
- Enter numbers in the input field
- If you enter incorrect characters, the input field will be highlighted in red
- click the "Find GCD and LOC" button
How to enter numbers
- Numbers are entered separated by a space, period or comma
- The length of entered numbers is not limited, so finding GCD and LCM of long numbers is not difficult
What are GCD and NOC?
Greatest common divisor several numbers is the largest natural integer by which all original numbers are divisible without a remainder. The greatest common divisor is abbreviated as GCD.
Least common multiple several numbers is the smallest number that is divisible by each of the original numbers without a remainder. The least common multiple is abbreviated as NOC.
How to check that a number is divisible by another number without a remainder?
To find out whether one number is divisible by another without a remainder, you can use some properties of divisibility of numbers. Then, by combining them, you can check the divisibility of some of them and their combinations.
Some signs of divisibility of numbers
1. Divisibility test for a number by 2
To determine whether a number is divisible by two (whether it is even), it is enough to look at the last digit of this number: if it is equal to 0, 2, 4, 6 or 8, then the number is even, which means it is divisible by 2.
Example: determine whether the number 34938 is divisible by 2.
Solution: We look at the last digit: 8 - that means the number is divisible by two.
2. Divisibility test for a number by 3
A number is divisible by 3 when the sum of its digits is divisible by three. Thus, to determine whether a number is divisible by 3, you need to calculate the sum of the digits and check whether it is divisible by 3. Even if the sum of the digits is very large, you can repeat the same process again.
Example: determine whether the number 34938 is divisible by 3.
Solution: We count the sum of the numbers: 3+4+9+3+8 = 27. 27 is divisible by 3, which means the number is divisible by three.
3. Divisibility test for a number by 5
A number is divisible by 5 when its last digit is zero or five.
Example: determine whether the number 34938 is divisible by 5.
Solution: look at the last digit: 8 means the number is NOT divisible by five.
4. Divisibility test for a number by 9
This sign is very similar to the sign of divisibility by three: a number is divisible by 9 when the sum of its digits is divisible by 9.
Example: determine whether the number 34938 is divisible by 9.
Solution: We count the sum of the numbers: 3+4+9+3+8 = 27. 27 is divisible by 9, which means the number is divisible by nine.
How to find GCD and LCM of two numbers
How to find the gcd of two numbers
Most in a simple way Calculating the greatest common divisor of two numbers is to find all possible divisors of these numbers and select the largest of them.
Let's consider this method using the example of finding GCD(28, 36):
- We factor both numbers: 28 = 1·2·2·7, 36 = 1·2·2·3·3
- We find common factors, that is, those that both numbers have: 1, 2 and 2.
- We calculate the product of these factors: 1 2 2 = 4 - this is the greatest common divisor of the numbers 28 and 36.
How to find the LCM of two numbers
There are two most common ways to find the least multiple of two numbers. The first method is that you can write down the first multiples of two numbers, and then choose among them a number that will be common to both numbers and at the same time the smallest. And the second is to find the gcd of these numbers. Let's consider only it.
To calculate the LCM, you need to calculate the product of the original numbers and then divide it by the previously found GCD. Let's find the LCM for the same numbers 28 and 36:
- Find the product of numbers 28 and 36: 28·36 = 1008
- GCD(28, 36), as already known, is equal to 4
- LCM(28, 36) = 1008 / 4 = 252 .
Finding GCD and LCM for several numbers
The greatest common divisor can be found for several numbers, not just two. To do this, the numbers to be found for the greatest common divisor are decomposed into prime factors, then the product of the common prime factors of these numbers is found. You can also use the following relation to find the gcd of several numbers: GCD(a, b, c) = GCD(GCD(a, b), c).
A similar relationship applies to the least common multiple: LCM(a, b, c) = LCM(LCM(a, b), c)
Example: find GCD and LCM for numbers 12, 32 and 36.
- First, let's factorize the numbers: 12 = 1·2·2·3, 32 = 1·2·2·2·2·2, 36 = 1·2·2·3·3.
- Let's find the common factors: 1, 2 and 2.
- Their product will give GCD: 1·2·2 = 4
- Now let’s find the LCM: to do this, let’s first find the LCM(12, 32): 12·32 / 4 = 96 .
- To find the NOC of everyone three numbers, you need to find GCD(96, 36): 96 = 1·2·2·2·2·2·3 , 36 = 1·2·2·3·3 , GCD = 1·2·2·3 = 12 .
- LCM(12, 32, 36) = 96·36 / 12 = 288.
Let's solve the problem. We have two types of cookies. Some are chocolate and others are plain. There are 48 chocolate ones, and 36 plain ones. You need to make the maximum possible number of gifts from these cookies, and you need to use them all.
First, let's write down all the divisors of each of these two numbers, since both of these numbers must be divisible by the number of gifts.
We get
- 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48.
- 36: 1, 2, 3, 4, 6, 9, 12, 18, 36.
Let us find among the common divisors that both the first and second numbers have.
Common factors will be: 1, 2, 3, 4, 6, 12.
The greatest common factor of all is the number 12. This number is called the greatest common factor of the numbers 36 and 48.
Based on the results obtained, we can conclude that 12 gifts can be made from all the cookies. One such gift will contain 4 chocolate cookies and 3 regular cookies.
Finding the Greatest Common Divisor
- Greatest natural number, by which two numbers a and b are divided without remainder, is called the greatest common divisor of these numbers.
Sometimes the abbreviation GCD is used to shorten the entry.
Some pairs of numbers have one as their greatest common divisor. Such numbers are called mutually prime numbers. For example, the numbers 24 and 35 have GCD =1.
How to find the greatest common divisor
In order to find the greatest common divisor, it is not necessary to write down all the divisors of the given numbers.
You can do it differently. First, factor both numbers into prime factors.
- 48 = 2*2*2*2*3,
- 36 = 2*2*3*3.
Now, from the factors that are included in the expansion of the first number, we will cross out all those that are not included in the expansion of the second number. In our case, these are two deuces.
- 48 = 2*2*2*2*3 ,
- 36 = 2*2*3 *3.
The factors remaining are 2, 2 and 3. Their product is 12. This number will be the greatest common divisor of the numbers 48 and 36.
This rule can be extended to the case of three, four, etc. numbers.
General scheme for finding the greatest common divisor
- 1. Divide numbers into prime factors.
- 2. From the factors included in the expansion of one of these numbers, cross out those that are not included in the expansion of other numbers.
- 3. Calculate the product of the remaining factors.
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Problems on GCD and LCM of numbers Work of a 6th grade student of the MCOU "Kamyshovskaya secondary school" Lantsinova Aisa Supervisor Zoya Erdnigoryaevna Goryaeva, mathematics teacher p. Kamyshevo, 2013
An example of finding the gcd of the numbers 50, 75 and 325. 1) Let's factor the numbers 50, 75 and 325 into prime factors. 50= 2 ∙ 5 ∙ 5 75= 3 ∙ 5 ∙ 5 325= 5 ∙ 5 ∙ 13 2) From the factors included in the expansion of one of these numbers, we cross out those that are not included in the expansion of the others. 50= 2 ∙ 5 ∙ 5 75= 3 ∙ 5 ∙ 5 325= 5 ∙ 5 ∙13 3) Find the product of the remaining factors 5 ∙ 5 = 25 Answer: GCD (50, 75 and 325) = 25 The largest natural number by which When numbers a and b are divided without remainder, the greatest common divisor of these numbers is called the greatest common divisor of these numbers.
An example of finding the LCM of the numbers 72, 99 and 117. 1) Let's factor the numbers 72, 99 and 117 into prime factors. 72 = 2 ∙ 2 ∙ 2 ∙ 3 ∙ 3 99 = 3 ∙ 3 ∙ 11 117 = 3 ∙ 3 ∙13 2) Write down the factors included in the expansion of one of the numbers 2 ∙ 2 ∙ 2 ∙ 3 ∙ 3 and add to them the missing factors of the remaining numbers. 2 ∙ 2 ∙ 2 ∙ 3 ∙ 3 ∙ 11 ∙ 13 3) Find the product of the resulting factors. 2 ∙ 2 ∙ 2 ∙ 3 ∙ 3 ∙ 11 ∙ 13= 10296 Answer: LCM (72, 99 and 117) = 10296 The least common multiple of natural numbers a and b is the smallest natural number that is a multiple of a and b.
The sheet of cardboard has the shape of a rectangle, the length of which is 48 cm and the width is 40 cm. This sheet must be cut into equal squares without waste. What are the largest squares that can be obtained from this worksheet and how many? Solution: 1) S = a ∙ b – area of the rectangle. S= 48 ∙ 40 = 1960 cm². – area of cardboard. 2) a – side of the square 48: a – the number of squares that can be laid along the length of the cardboard. 40: a – the number of squares that can be laid across the width of the cardboard. 3) GCD (40 and 48) = 8 (cm) – side of the square. 4) S = a² – area of one square. S = 8² = 64 (cm²) – area of one square. 5) 1960: 64 = 30 (number of squares). Answer: 30 squares with a side of 8 cm each. GCD problems
The fireplace in the room must be tiled in the shape of a square. How many tiles will be needed for a fireplace measuring 195 ͯ 156 cm and what are they? largest dimensions tiles? Solution: 1) S = 196 ͯ 156 = 30420 (cm²) – S of the fireplace surface. 2) GCD (195 and 156) = 39 (cm) – side of the tile. 3) S = a² = 39² = 1521 (cm²) – area of 1 tile. 4) 30420: = 20 (pieces). Answer: 20 tiles measuring 39 ͯ 39 (cm). GCD problems
A garden plot measuring 54 ͯ 48 m around the perimeter must be fenced; to do this, concrete pillars must be placed at regular intervals. How many poles need to be brought for the site, and at what maximum distance from each other will the poles be placed? Solution: 1) P = 2(a + b) – perimeter of the site. P = 2(54 + 48) = 204 m. 2) GCD (54 and 48) = 6 (m) – the distance between the pillars. 3) 204: 6 = 34 (pillars). Answer: 34 pillars, at a distance of 6 m. GCD problems
Bouquets were collected from 210 burgundy, 126 white, and 294 red roses, with each bouquet containing an equal number of roses of the same color. Which greatest number bouquets were made from these roses and how many roses of each color are in one bouquet? Solution: 1) GCD (210, 126 and 294) = 42 (bouquets). 2) 210: 42 = 5 (burgundy roses). 3) 126: 42 = 3 (white roses). 4) 294: 42 = 7 (red roses). Answer: 42 bouquets: 5 burgundy, 3 white, 7 red roses in each bouquet. GCD problems
Tanya and Masha bought the same number of postal kits. Tanya paid 90 rubles, and Masha paid 5 rubles. more. How much does one set cost? How many sets did each person buy? Solution: 1) 90 + 5 = 95 (rub.) Masha paid. 2) GCD (90 and 95) = 5 (rub.) – price of 1 set. 3) 980: 5 = 18 (sets) – bought by Tanya. 4) 95: 5 = 19 (sets) – bought by Masha. Answer: 5 rubles, 18 sets, 19 sets. GCD problems
Three tourist boat trips begin in the port city, the first of which lasts 15 days, the second – 20 and the third – 12 days. Having returned to the port, the ships set off again on the same day. Today, ships left the port on all three routes. In how many days will they go sailing together again for the first time? How many trips will each ship make? Solution: 1) NOC (15,20 and 12) = 60 (days) – meeting time. 2) 60: 15 = 4 (voyages) – 1 ship. 3) 60: 20 = 3 (voyages) – 2 ships. 4) 60: 12 = 5 (flights) – 3 ships. Answer: 60 days, 4 flights, 3 flights, 5 flights. NOC tasks
Masha bought eggs for the Bear at the store. On the way to the forest, she realized that the number of eggs is divisible by 2,3,5,10 and 15. How many eggs did Masha buy? Solution: LOC (2;3;5;10;15) = 30 (eggs) Answer: Masha bought 30 eggs. NOC tasks
It is required to make a box with a square bottom to accommodate boxes measuring 16 ͯ 20 cm. What is the shortest length of the side of the square bottom to fit the boxes tightly into the box? Solution: 1) LCM (16 and 20) = 80 (boxes). 2) S = a ∙ b – area of 1 box. S = 16 ∙ 20 = 320 (cm²) – bottom area of 1 box. 3) 320 ∙ 80 = 25600 (cm²) – area of the square bottom. 4) S = a² = a ∙ a 25600 = 160 ∙ 160 – dimensions of the box. Answer: 160 cm is the side of the square bottom. NOC tasks
Along the road from point K there are power poles every 45 m. They decided to replace these poles with others, placing them at a distance of 60 m from each other. How many pillars were there and how many will there be? Solution: 1) LCM (45 and 60) = 180. 2) 180: 45 = 4 – there were pillars. 3) 180: 60 = 3 – became pillars. Answer: 4 pillars, 3 pillars. NOC tasks
How many soldiers are marching on the parade ground if they march in formation of 12 people in a line and change into a column of 18 people in a line? Solution: 1) NOC (12 and 18) = 36 (people) - marching. Answer: 36 people. NOC tasks
Finding the greatest common divisor of three or more numbers can be reduced to sequentially finding the gcd of two numbers. We mentioned this when studying the properties of GCD. There we formulated and proved the theorem: the greatest common divisor of several numbers a 1 , a 2 , …, a k equal to the number dk, which is found by sequential calculation GCD(a 1 , a 2)=d 2, GCD(d 2 , a 3)=d 3, GCD(d 3 , a 4)=d 4, …,GCD(d k-1 , a k)=d k.
Let's see what the process of finding the gcd of several numbers looks like by looking at the solution to the example.
Example.
Find the greatest common divisor of four numbers 78 , 294 , 570 And 36 .
Solution.
In this example a 1 =78, a 2 =294, a 3 =570, a 4 =36.
First, using the Euclidean algorithm, we determine the greatest common divisor d 2 first two numbers 78 And 294 . When dividing we get the equalities 294=78 3+60; 78=60 1+18;60=18·3+6 And 18=6·3. Thus, d 2 =GCD(78, 294)=6.
Now let's calculate d 3 =GCD(d 2, a 3)=GCD(6, 570). Let's use the Euclidean algorithm again: 570=6·95, hence, d 3 =GCD(6, 570)=6.
It remains to calculate d 4 =GCD(d 3, a 4)=GCD(6, 36). Because 36 divided by 6 , That d 4 =GCD(6, 36)=6.
Thus, the greatest common divisor of the four given numbers is equal to d 4 =6, that is, GCD(78, 294, 570, 36)=6.
Answer:
GCD(78, 294, 570, 36)=6.
Factoring numbers into prime factors also allows you to calculate the gcd of three or more numbers. In this case, the greatest common divisor is found as the product of all common prime factors of the given numbers.
Example.
Calculate the gcd of the numbers from the previous example using their prime factorizations.
Solution.
Let's break down the numbers 78 , 294 , 570 And 36 by prime factors, we get 78=2·3·13,294=2·3·7·7, 570=2 3 5 19, 36=2·2·3·3. The common prime factors of all given four numbers are the numbers 2 And 3 . Hence, GCD(78, 294, 570, 36)=2·3=6.
Answer:
GCD(78, 294, 570, 36)=6.
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Finding GCD of Negative Numbers
If one, several, or all of the numbers whose greatest divisor is to be found are negative numbers, then their gcd is equal to the greatest common divisor of the moduli of these numbers. This is due to the fact that opposite numbers a And −a have the same divisors, as we discussed when studying the properties of divisibility.
Example.
Find the gcd of negative integers −231 And −140 .
Solution.
The absolute value of a number −231 equals 231 , and the modulus of the number −140 equals 140 , And GCD(−231, −140)=GCD(231, 140). The Euclidean algorithm gives us the following equalities: 231=140 1+91; 140=91 1+49; 91=49 1+42; 49=42 1+7 And 42=7 6. Hence, GCD(231, 140)=7. Then the desired greatest common divisor of negative numbers is −231 And −140 equals 7 .
Answer:
GCD(−231, −140)=7.
Example.
Determine the gcd of three numbers −585 , 81 And −189 .
Solution.
When finding the greatest common divisor, negative numbers can be replaced by them absolute values, that is, GCD(−585, 81, −189)=GCD(585, 81, 189). Number expansions 585 , 81 And 189 into prime factors have the form 585=3·3·5·13, 81=3·3·3·3 And 189=3·3·3·7. The common prime factors of these three numbers are 3 And 3 . Then GCD(585, 81, 189)=3·3=9, hence, GCD(−585, 81, −189)=9.
Answer:
GCD(−585, 81, −189)=9.
35. Roots of a polynomial. Bezout's theorem. (33 and above)
36. Multiple roots, criterion for multiplicity of roots.