The concepts of multiples and dividers of a natural number extend to the set of whole numbers. When dealing with the subject of multiples and divisors, we refer to numerical sets that satisfy some conditions. Multiples are found after multiplication by whole numbers, and divisors are numbers divisible by a certain number.
Because of this, we will find subsets of the integers, as the elements of the sets of multiples and divisors are elements of the set of integers. To understand what prime numbers are, it is necessary to understand the concept of divisors.
multiples of a number
be The and B two known integers, the number The is multiple of B if and only if there is an integer k such that The = B · k. Thus, the set of multiples in Theis obtained by multiplyingThefor all whole numbers, the results of these multiplications are the multiples of The.
For example, let's list the first 12 multiples of 2. For this we have to multiply the number 2 by the first 12 whole numbers, like this:
2 · 1 = 2
2 · 2 = 4
2 · 3 = 6
2 · 4 = 8
2 · 5 = 10
2 · 6 = 12
2 · 7 = 14
2 · 8 = 16
2 · 9 = 18
2 · 10 = 20
2 · 11 = 22
2 · 12 = 24
Therefore, multiples of 2 are:
M(2) = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24}
Note that we only listed the first 12 numbers, but we could have listed as many as needed, as the list of multiples is given by multiplying a number by all the integers. Thus, the set of multiples is infinite.
To check whether or not a number is a multiple of another, we must find a whole number so that the multiplication between them results in the first number. See the examples:
→ The number 49 is a multiple of 7, because there is an integer that, multiplied by 7, results in 49.
49 = 7 · 7
→ The number 324 is a multiple of 3, as there is an integer that, multiplied by 3, results in 324.
324 = 3 · 108
→ The number 523 no is a multiple of 2 because there is no integer which, multiplied by 2, results in 523.
523 = 2 · ?
Read too: Properties of multiplication that facilitate mental calculation
Multiples of 4
As we have seen, to determine the multiples of the number 4, we must multiply the number 4 by whole numbers. Thus:
4 · 1 = 4
4 · 2 = 8
4 · 3 = 12
4 · 4 = 16
4 · 5 = 20
4 · 6 = 24
4 · 7 = 28
4 · 8 = 32
4 · 9 = 36
4 · 10 = 40
4 · 11 = 44
4 · 12 = 48
...
Therefore, multiples of 4 are:
M(4) = {4, 8, 12, 16, 20. 24, 28, 32, 36, 40, 44, 48, … }
Multiples of 5
Analogously, we have multiples of 5.
5 · 1 = 5
5 · 2 = 5
5 · 3 = 15
5 · 4 = 20
5 · 5 = 25
5 · 6 = 30
5 · 7 = 35
...
Hence, the multiples of 5 are: M(5) = {5, 10, 15, 20, 25, 30, 35, 40, 45, … }
one number dividers
be The and B two known integers, let's say B is divider of The if the number B is multiple of The, that is, the division in between B and The is exact (must leave rest 0).
See some examples:
→ 22 is a multiple of 2, so 2 is a divisor of 22.
→ 63 is a multiple of 3, so 3 is a divisor of 63.
→ 121 is not a multiple of 10, so 10 is not a divisor of 121.
To list the divisors of a number, we must look for the numbers that divide it. Look:
– List the dividers of 2, 3 and 20.
D(2) = {1, 2}
D(3) = {1,3}
D(20) = {1, 2, 4, 5, 10, 20}
Note that numbers in the list of divisors are always divisible by the number in question and that the highest value that appears in this list is the number itself., since no number greater than it will be divisible by it.
For example, in divisors of 30, the largest value in this list is 30 itself, as no number greater than 30 will be divisible by it. Thus:
D(30) = {1, 2, 3, 5, 6, 10, 15, 30}
Know more: Fun Facts About Dividing Natural Numbers
Ownership of multiples and divisors
These properties are related to division between two integers. Note that when an integer is a multiple of another, it is also divisible by that other number.
Consider the division algorithm so that we can better understand the properties.
N = d · q + r, where q and r are integers.
remember that N is called of dividend;d, for divider;q, for quotient; and r, by the way.
→ Property 1: The difference between the dividend and the remainder (N – r) is a multiple of the divisor, or the number d is a divisor of (N – r).
→ Property 2: (N – r + d) is a multiple of d, that is, the number d is a divisor of (N – r + d).
See the example:
– When performing the division of 525 by 8, we obtain quotient q = 65 and remainder r = 5. Thus, we have the dividend N = 525 and the divisor d = 8. See that the properties are satisfied, since (525 – 5 + 8) = 528 is divisible by 8 and:
528 = 8 · 66
Prime numbers
You Prime numbers are those that have as a divisor in their listing only the number 1 and the number itself. To check whether a number is prime or not, one of the most trivial methods is to list the divisors of that number. If numbers more than 1 and the number in question appear, it is not prime.
→ Check which are the prime numbers between 2 and 20. For that, let's list the divisors of all these numbers between 2 and 20.
D(2) = {1, 2}
D(3) = {1,3}
D(4) = {1, 2, 4}
D(5) = {1, 5}
D(6) = {1, 2, 3, 6}
D(7) = {1, 7}
D(8) = {1, 2, 4, 8}
D(9) = {1, 3, 9}
D(10) = {1, 2, 5, 10}
D(11) = {1, 11}
D(12) = {1, 2, 3, 4, 6, 12}
D(13) = {1, 13}
D(14) = {1, 2, 7, 14}
D(15) = {1, 3, 5, 15}
D(16) = {1, 2, 4, 16}
D(17) = {1, 17}
D(18) = {1, 2, 3, 6, 9, 18}
D(19) = {1, 19}
D(20) = {1, 2, 4, 5, 10, 20}
So the prime numbers between 2 and 20 are:
{2, 3, 5, 7, 11, 13, 17 and 19}
Note that the set is from some of the first primes, this list goes on. Note that the larger the number, the harder it becomes to tell if it is prime or not.
Read more: Irrational numbers: those that cannot be represented in fractions
solved exercises
question 1 – (UMC-SP) The number of elements in the set of prime divisors of 60 is:
a) 3
b) 4
c) 5
d) 10
Solution
Alternative A
We will first list the divisors of 60 and then look at which ones are prime.
D(60) = {1, 2, 3, 5, 6, 10, 12, 15, 20, 30, 60}
Of these numbers, we have that are prime:
{2, 3, 5}
Therefore, the number of prime divisors of 60 is 3.
question 2 – Write all natural numbers less than 100 and multiples of 15.
Solution
We know that the multiples of 15 are the results of multiplying the number 15 by all the integers. Since the exercise asks to write the natural numbers less than 100 and which are multiples of 15, we must multiply 15 by all numbers greater than zero, until we find the largest multiple before 100, thus:
15 · 1 = 15
15 · 2 = 30
15 · 3 = 45
15 · 4 = 60
15 · 5 = 75
15 · 6 = 90
15 · 7 = 105
Therefore, natural numbers less than 100 and multiples of 15 are:
{15, 30, 45, 60, 75, 90}
question 3 – What is the largest multiple of 5 between 100 and 1001?
Solution
To determine the largest multiple of 5 between 100 and 1001, simply identify the first multiple of 5 back to front.
1001 is not a multiple of 5, as there is no integer that, multiplied by 5, results in 1001.
1000 is a multiple of 5, since 1000 = 5 · 200.
Therefore, the largest multiple of 5, between 100 and 1001, is 1000.
by Robson Luiz
Maths teacher
Source: Brazil School - https://brasilescola.uol.com.br/matematica/multiplos-divisores.htm