Introduction to Factorization

8/12/2019 Introduction to Factorization

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F 2014 by nielsMathTutorials

Introduction to

FactorizationThe Greatest Common Divisor (GCD), also known as the Highest Common Factor (HCF),

is the biggest number that can divide two or more numbers.

In this tutorial, we will consider the concept of factors, or divisors. We will then discuss

prime numbers, and how to express a number that is not a prime number as a product of

prime numbers.

Finally, we will consider the concept of common factors, and most importantly, how to

find the HCF of two or more numbers.

However, in order to gain mastery of the methods presented here, practice questions are

provided at the end of every section.

Contents

1. Factors 12. Factorization 33. Prime Numbers 74. Prime Factors 105. Prime Factorization 136. Common Factors 187. Highest Common Factors 19

Answers to Exercise Questions 22

1. Factors

Division of one whole number by another is a very familiar arithmetic operation. Whendividing one whole number by another whole number, another whole number could be

the result.

For instance, 5 divides 40 eight times exactly. (Or, 40 = 5 8)

Also, 4 divides 40 ten times exactly. (Or, 40 = 4 10)


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Since 5 and 4 divide 40 exactly, without leaving remainders, both of them are called

Factors, or Divisors,of 40.

However, 6 is not a factor of 40 because

40 6 = 6 times and 4 remainders

For the same reason, 11 is also not a factor of 40, because

40 11 = 3 times and 7 remainders

Definition

Letx stand for any positive whole number

A Factorof xdivides xwithout leaving a remainder.

Note

1 can divide every other number without leaving remainders. So 1

is a factor of every number.

Note

Since every number can divide itself exactly, every number is a

factor of itself.

The examples below show how to determine if a number is a factor of another

Example

Say whether 3, 4, and 5 are factors of 24. Give reasons.3 is a factor of 24

Reason: 24contains3eight times and 0remainder

OR, 24 = 3 8

4 is a factor of 24

Reason: 24contains 4six times and 0remainder

OR, 24 = 4 6

5 is not a factor of 24Reason: 24contains 5four times and 4remainders

OR, 24 = (5 4) + 4


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Example

Say whether 10, 14, and 16 are factors of 60. Give reasons.

10 is a factor of 60

Reason: 60contains 10six times and 0remainderOR, 60 = 10 6

14 is not a factor of 60

Reason: 60contains 14four times and 4remainders

OR, 60 = (14 4) + 4


Reason: 60contains 16three times and 12remainders

OR, 60 = (16 3) + 12

Exercise 1

1. Which of these are factors of 75?(i) 2 (ii) 3 (iii) 5 (iv) 15

2. Which of these are NOT factors of 64?(i) 12 (ii) 8 (iii) 18 (iv) 16

3. Say whether 7, 9 and 15 are factors of 45. Give reasons.4. Say whether 10, 12 and 14 are factors of 70. Give reasons.

2. Factorization

It is possible to write a positive whole number, say 40, as a product of its factors.

For instance, 40 = 8 5 or 5 8

Notice that 5 divides 40 eight times, and 8 divides 40 five times.

Writing 40 as 5 8 is called Factoring, or Factorizing, 40 because we have now written

40 as a multiplication of two of its factors.

40 can also be factored, or factorized, in other ways, like

40 = 1 40 or 40 1

40 = 4 10 or 10 4

e.t.c.


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Definition

To Factorizea whole number means to write that number as a

product, or a multiplication, of its factors.

Note

Saying Factorizea number and Factora number both mean

the same thing.

The examples below show how to determine all the factors of a given number.

Example

Find all the factors of 16.

First find all the possible ways you can express 16 as a product of a pair of its

factors1 is a factor of every number, so

16 = 1 16

Next try 2: 2 is also a factor of 16, 16 contains 2 eight times

16 = 1 16

2 8

Next try 3: 3 is not a factor of 16Next try 4: 4 is a factor of 16, 16 contains 4 four times

16 = 1 16

2 8

4 4

Next try 5, then 6, then 7, until you get to 16

16 = 1 16

2 8

4 4

8 2 (Same as 2 8)

16 1 (Same as 1 16)


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Next try 5 5 is a factor 60 contains 5 twelve times

60 = 1 60

2 30

3 204 15

5 12

Next try 6 6 is also a factor 60 contains 6 ten times

60 = 1 60

2 30

3 20

4 15

5 12

6 10

Between 6 and 10, there is no factor of 60.

The next factor of 60 is 10, which is already there.

After 10, the next factor is 12. You can see that 12 is already there.

So we finished all the possibilities.

60 = 1 60

2 30

3 20

4 15

5 12

6 10

The factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60.

Exercise 2

Find all the factors of

(a)24 (b) 36 (c) 54 (d) 100

ollow the arrow


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3. Prime Numbers

Let us list the numbers from 1 to 15, along with their factors

Number Factors Number Factors1 1 9 1, 3, 9

2 1, 2 10 1, 2, 5, 10

3 1, 3 11 1, 11

4 1, 2, 4 12 1, 2, 3, 4, 6, 12

5 1, 5 13 1, 13

6 1, 2, 3, 6 14 1, 2, 7, 14

7 1, 7 15 1, 3, 5, 15

8 1, 2, 4, 8

From the above table, we can see that some numbers have only two factors: themselves

and 1. This simply means that the only numbers that can divide them are themselves and

1.

The first two of such numbers are 2 and 3. Can you locate the others?

Right! The rest are 5, 7, 11 and 13.

Such numbers are called Prime Numbers

Definition

A Prime Numberis a number that has only two factors, itself

and 1.

Note

1 is not a prime number. This is because it has only one factor;

itself.

From the table, we can see that we have 6 prime numbers between 1 and 15: 2, 3, 5, 7, 11

and 13.

If we wanted to get the prime numbers, say up to 100, it would be tedious to get all

factors for each number, check if they are only two factors for the number, and conclude

that the number is prime number.


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However, there is another more convenient method, called the Sieve of Eratosthenes,

which is described below:

List the whole numbers from 1 to the number you want to stop at, say 70, or maybe100

Strike out 1, since it has only one factor: itself. So it is not prime 2 is a prime. Leave it. Strike out every multiple of 2 beginning from 4. (4 = 2

2)

The first number not struck out by 2 is 3. So 3 is a prime. Leave it. Strike out everymultiple of 3 beginning from 9. (9 = 3 3)

The first number not struck out by 3 is 5. So 5 is also a prime. Leave it. Strike outevery multiple of 5 beginning from 25. (25 = 5 5)

The first number not struck out by 5 is 7. So 7 is a prime. Leave it. Strike out everymultiple of 7 beginning from 49. (49 = 7 7)

And so on until there are no more numbers to strike out. The remaining numbers are the

prime numbers in the range of numbers you chose.

The example below demonstrates how to use this method to find all the prime numbers

between 1 and 50.

Example

Use the Sieve of Eratosthenes method of finding prime numbers to find all primenumbers between 1 and 50.

List the numbers from 1 to 50

1 2 3 4 5 6 7 8 9 10

11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30

31 32 33 34 35 36 37 38 39 40

41 42 43 44 45 46 47 48 49 50


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Strike out 1; 1 is not a prime number. (Each number I strike out will be shaded)

1 2 3 4 5 6 7 8 9 10

11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 3031 32 33 34 35 36 37 38 39 40

41 42 43 44 45 46 47 48 49 50

The first prime number is 2. Leave it. Strike out every multiple of 2 starting with 2

2 = 4

1 2 3 4 5 6 7 8 9 10

11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 3031 32 33 34 35 36 37 38 39 40

41 42 43 44 45 46 47 48 49 50

3 is the first whole number that is not a multiple of 2. So it is prime, leave it. But

strike out every multiple of 3 starting with 3 3 = 9

1 2 3 4 5 6 7 8 9 10

11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 3031 32 33 34 35 36 37 38 39 40

41 42 43 44 45 46 47 48 49 50

5 is the first whole number that was not struck out by 3. So 5 is prime. Strike out

every multiple of 5 starting with 5 5 = 25

1 2 3 4 5 6 7 8 9 10

11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 3031 32 33 34 35 36 37 38 39 40

41 42 43 44 45 46 47 48 49 50

The next prime number is 7 (2, 3 or 5 did not succeed in striking it out)


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Strike out all multiples of 7 starting from 7 7 = 49

1 2 3 4 5 6 7 8 9 10

11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 3031 32 33 34 35 36 37 38 39 40

41 42 43 44 45 46 47 48 49 50

The next prime number is 11, (For the same reasons as before). 11 11 = 121 is not in

the grid. So we can stop.

The remaining numbers are the prime numbers between 1 and 50:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, and 47.

Exercise 3

1. In the introduction, the only prime numbers we had were 2, 3, 5, 7, 11 and 13. Theexample above produced prime numbers up to 47. Try to show that the extra

numbers produced are prime numbers by finding their factors.

2. Find all the prime numbers between 1 and 100.4. Prime Factors

If we were to list all the factors of 28, we would get the following six factors:

28 = 1 28

2 14

4 7

That is: 1, 2, 4, 7, 14, 28

Notice that out of the six factors of 28, 2 and 7 are prime numbers. These are the primefactorsof 28.


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Also, listing the eight factors of 30, we get

30 = 1 30

2 15

3 10

5 6

That is: 1, 2, 3, 5, 6, 10, 15, 30

Here there are also prime numbers: 2, 3 and 5. These are the prime factorsof 30

Definition

Letx stand for any positive whole number

A Prime Factorof xis a prime number which is also a factor of x.

Note

Not all factors of a number are prime numbers. However, if a

factor is also a prime number, it is called a prime factor.

Here are some examples

Example

Find the prime factors of 42.

First Work out the factors of 42

Working:

The factors of 42: 1, 2, 3, 6, 7, 14, 21 and 42.

Then get those factors which are prime numbers

Prime factors of 42: 2, 3, 7

42 = 1 42

2 21

3 14

6 7


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But a natural question might arise:

If I were to find all the factors of 450, I get more than one product

450 = 1 450

2 2253 150

5 90

6 75

9 50

10 45

15 30

18 25

Is there also more than one way to write 450 as a product of its prime factors?

To find out, let us carry out a little experiment

450 = 45 10 = (5 9) (2 5) = 5 (3 3) 2 5 = 5 3 3 2 5 450 = 9 50 = (3 3) (5 10) = 3 3 5 (2 5) = 3 3 5 2 5 450 = 18 25 = (2 9) (5 5) = 2 (3 3) 5 5 = 2 3 3 5 5 450 = 3 150 = 3 10 15 = 3 (2 5) (3 5) = 3 2 5 3 5

Notice that no matter which product we started with, the final product of prime factors isalways a rearrangement of the product 2 3 3 5 5

This is actually a result in Mathematics called The Fundamental Theorem of

Arithmetic.

The Fundamental Theorem of Arithmetic

Stated informally; if a whole number xis not a prime number, we

can always write it as a product of prime numbers in only one way

(if we ignore any rearrangement of the factors). The product

would contain only the prime factors of x.

The following examples will illustrate this.


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Example

Express 180 as a product of its prime factors.

180 = 2 90

= 2 2 45= 2 2 3 15

= 2 2 3 3 5

Example


550 = 2 275

= 2 5 55

= 2 5 5 11

Alternatively, we could use short division repeatedly to solve each of the above problems.

This method is also called Trial Division.

Trial Division requires knowledge of the prime numbers.

The next set of examples shows how to use trial division to arrive at a prime factorization

of whole numbers.

ExampleWrite 180 as a product of its prime factors using trial division.

Try dividing 180 by the first prime number, 2. Since 2 is a prime factor of 180, we

have

2 180

90

Try dividing 90 by 2 again

2 180

2 90

45


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2 is not a factor of 45. Then try 45 divided by the next prime number 3. Since 3 is a

prime factor of 45, we have

2 180

2 90

3 45

15

Now we have 15. If we try 3 again, 3 can still divide 15.

2 180

2 90

3 45

3 15

5

The only number that can divide 5 is 5. (Remember, 5 is a prime number) So divide

by 5 to complete the process

2 180

2 90

3 45

3 155 5

1

Example


For 550, the only prime numbers that will divide are 2, 5, and 11 (the prime factors

of 550)

First try 2:

2 550

275

2 2 3 3 5 = 180


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Try 5:

2 550

5 275

55Try 5 again:

2 550

5 275

5 55

11

Finally, try 11:

2 550

5 275

5 55

11 11

1

Example

Use Trial division to find the prime factorization of 600.

600 = 2 2 2 3 5 5

2 600

2 300

2 150

3 75

5 25

5 5

1

2 5 5 11 = 550


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Example

Use Trial division to find the prime factorization of 588.

588 = 2 2 3 7 7

Exercise 5

Write the following numbers as a product of their respective prime factors(b)20 (b) 30 (c) 12 (d) 66 (e) 78

6. Common Factors

In the previous sections, we discussed how to find the factors of a single whole number.

So, if a number divides x without leaving a remainder, that number is a factor of x. For

example, 1, 2, 3, 4, 6 and 12 are all factors of 12 because they divide 12 without leaving a

remainder.

Now to something slightly different: Factors common to two or more numbers. Say, you

are given two numbers 12 and 18. Of course 1 can divide both 12 and 18. So 1 is a common

factor of 12 and 18. We also have 2 as a common factor of 12 and 18. Actually, all the

common factors of 12 and 18 are 1, 2, 3, and 6.

Definition

A Common Factorof two or more numbers divides the two

numbers without leaving any remainders.

The following examples show how to find all the Common factors of two or more

numbers.

2 588

2 294

3 147

7 49

7 7

1


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Example

Get all the common factors of 18 and 24.

Factors of 18: 1, 2,3, 6, 9, 18

Factors of 24: 1, 2,3, 4, 6, 8, 12, 24Common factors of 16 and 18: 1, 2, 3, 6

Example

Get all the common factors of 16, 24, and 40.

Factors of 16: 1, 2, 4, 8, 16

Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24

Factors of 40: 1, 2, 4, 5, 8, 10, 20, 40

Common factors of 16, 24, 40: 1, 2, 4, 8

Exercise 6

List all the common factors of the following

(a) 18 and 27 (b) 12 and 30 (c) 15, 24, and 33

7. Highest Common Factors

Consider the following example

Factors of 30: 1, 2, 3, 5, 6, 10, 15, 30

Factors of 45: 1, 3, 5, 9, 15, 45

Common factors of 30 and 45: 1, 3, 5, 15

Four factors are common to 30 and 45. But 15 is the largest number that can divide both

30 and 45. So 15 is said to be the highest common factor of 30 and 45.

Definition

The Highest Common Factorof two numbers is the largest

whole number that is a common factor of the two numbers.

But the above method for finding the HCF of 30 and 45 is a little bit tedious. It is possible

to use the prime factorization method to find the HCF of 30 and 45.

The following examples will illustrate this.


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Example

Use the prime factorization method to find the HCF of 30 and 45

In this method, first express 30 and 45 each as products of their prime factors. Next,

carefully observe which smaller product in 30 is the same with a smaller product in45. The smaller product will be equal to the HCF of 30 and 45.

2 30

3 15

5 5

1

3 45

3 15

5 5

1

Example

Find the HCF of 16 and 28.

16 = 2 2 2 2

28 = 2 2 7

HCF = 2 2 = 4

Example

Find the HCF of 12, 18, and 30.

12 = 2 2 3

18 = 2 3 3

30 = 2 3 5

HCF = 2 3 = 6

HCF = 3 5 = 15

30 = 2 3 5

45 = 3 3 5


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Example

Find the HCF of 20 and 21.

20 = 2 2 5

21 = 3 7HCF = (Hmnh nothing) just say 1

Why?

Lets verify using the longmethod

Factors of 20: 1, 2, 4, 5, 10, 20

Factors of 21: 1, 3, 7, 21

Common factors of 20 and 21: 1

The HCF: 1

Exercise 7

Find the HCF of the following(a)28 and 70 (b) 24 and 36 (c) 27 and 90 (d) 156, 117 and 195


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Answers to Exercise Questions

Exercise 1

(a)3, 5 and 15 are factors of 75 because75 =32575 = 515

75 = 155

(b)12 is NOT a factor of 64 because it leaves 4 remainders when dividing 64.18 is also NOT a factor of 64 because it leaves 10 remainders after

dividing 64.

(c)7 is not a factor of 45Reason: 45contains 7six times and3remainder

OR, 45 = (7 6) + 3

9 is a factor of 45

Reason: 45contains 9five times and 0remainders

OR, 45 = 9 5


Reason: 45contains 15three times and 0remainders

OR, 45 = 15 3

(d)10 is a factor of 70Reason: 70contains 10seven times and 0remainder

OR, 70 = 10 7



OR, 70 = (12 5) + 10



OR, 70 = 14 5


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Exercise 2

(a)Factors of 24 = 1, 2, 3, 4, 6, 8, 12 and 24(b)Factors of 36 = 1, 2, 3, 4, 6, 9, 12, 18 and 36(c)Factors of 54 = 1, 2, 3, 6, 9, 18, 27 and 54(d)Factors of 100 = 1, 2, 4, 5, 10, 20, 25, 50 and 100

Exercise 3

(2) 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97.Exercise 4

(a)Prime Factors of 20 = 2 and 5.(b)Prime Factors of 37 = 37 only.(c)Prime Factors of 50 = 2 and 5.(d)Prime Factors of 91 = 7 and 13.(e)Prime Factors of 120 = 2, 3 and 5.

Exercise 5

(a)20 = 2 2 5(b)30 = 2 3 5(c)12 = 2 2 3(d)66 = 2 3 11(e)78 = 2 3 13

Exercise 6

(a)Common Factors of 18 and 27 = 1, 3 and 9(b)Common Factors of 12 and 30 = 1, 2, 3 and 6(c)Common Factors of 15, 24 and 33 = 1 and 3

Exercise 7

(a)HCF of 28 and 70 = 2 7 = 14(b)HCF of 24 and 36 = 2 2 3 = 12(c)HCF of 27 and 90 = 3 3 = 9(d)HCF of 156, 117 and 195 = 3 13 = 39

Introduction to Factorization

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