Prime Factorization and Division Method For HCF

Introduction

The prime factorization and highest common factor (HCF) are two fundamental concepts in arithmetic. Factorization includes the split of an integer into several equal parts. They are extensively used for exchanging money, comparing prices, making arithmetic calculations, optimizing resources, etc. In this tutorial, we will learn about the HCF, methods to find HCF, prime factorization, and repeated division formula with solved examples.

Factors

The factor is defined as an integer when it divides a number; the remained results in zero. In other words, if the product of two integers results in a third number, then the two integers are said to be the factors of the resultant number. For example, 12 can be factorized as below.

$$\mathrm{1\times\:12\:=\:12}$$

$$\mathrm{12\times\:1\:=\:12}$$

$$\mathrm{3\times\:4\:=\:12}$$

$$\mathrm{2\times\:6\:=\:12}$$

Therefore, factors of 12 are 1, 2, 3, 4, 6, and 12.

There are two methods used to find the factors of a number.

• Multiplication method

• Division method

HCF

The highest common factor(HCF) of two or more two numbers is defined as the highest factor of both numbers. In other words, it is defined as the highest integer that divides the numbers completely. It is also known as the greatest common divisor (GCD). For example, consider two numbers 10 and 15.

The factors of 10 are 1, 2, 5, and 10.

The factors of 15 are 1, 3, 5, and 15.

Among the factors, 1 and 5 are the common factors of the above numbers. However, 5 is the highest common divisor or factor. Therefore, the HCF of 10 and 15 is 5.

Methods to find HCF

Three methods are used in mathematics to determine the HCF of numbers.

• Prime factorization

• Division method

• Listing factors

In this tutorial, we will focus on the prime factorization and division method the evaluate the HCF of integers.

Prime Factorization

Prime factorization is a way of representing large numbers in terms of the product of prime numbers. In other words, the breakdown of the number in terms of prime numbers (i.e., 2, 3, 5, 7, 11, etc.) is called prime factorization. Let’s consider that 260 is a composite number. The prime factorization of 260 can be represented as

$$\mathrm{260\:=\:2\times\:2\times\:5\times\:13}$$

In addition, there are two methods for prime factorization.

Factor tree method

We have to follow the below steps to find the prime factorization of a number.

• Write the given number on the top of the tree

• Write the pair of factors as the branches of a tree.

• Again, split the factors obtained in the above step.

• Repeat the above steps until we get the prime factors that cannot be factorized further.

Division method

We have to follow the below steps to find the prime factorization of a number.

• Divide the given number by the smallest prime number to get the remainder as zero

• Divide the quotient by the smallest prime number.

• Repeat the above step until the quotient becomes 1.

• The obtained divisors are the required prime factors of the number

Repeated Division or Euclid's Division Lemma

Euclid’s division lemma (lemma means theorem) states that a non-zero integer exists in such a way that if the number is divided by the integer, get a quotient and a remainder (non-zero).

Let’s consider the number ‘m’, and the integer ‘l’. Euclid’s division lemma can be mathematically expressed as

$$\mathrm{m\:=\:lr\:+\:s\:,\:where\:0\:\leq\:s\:\leq\:1}$$

Here ‘l’ is the quotient, and ‘s’ is the remainder. In addition, ‘m’ and ‘r’ are known as dividend and divisor, respectively.

Therefore, another way of representing Euclid’s Division lemma is

$$\mathrm{Dividend\:=\:(Divisor\:\times\:Quotient)\:+\:Remainder}$$

Now, we will see the proof of Euclid’s Division lemma.

Proof of Euclid’s Division lemma −

According to this theorem,

$$\mathrm{m\:=\:lr\:+\:s\:,\:where\:0\:\leq\:s\:\leq\:1}$$

Let’s consider m = 5 and l = 1

Therefore, $\mathrm{5\:=\:1\times\:5\:\div\:0}$

Here,$\mathrm{r\:=\:5\:and\:s\:=\:0}$

We can see that $\mathrm{0\leq\:0\leq\:1\:(0\leq\:s\leq\:1)}$

Now, consider m = 5 and l = 2

Therefore,$\mathrm{5\:=\:2\times\:2\:\div\:1}$

Here, r = 2 and s =1

We can see that $\mathrm{0\leq\:1\leq\:2\:(0\leq\:s\leq\:1)}$

Now, consider m = 5 and l = 3

Therefore, $\mathrm{5\:=\:3\times\:1\div\:2}$

Here, r = 1 and s = 2

We can see that $\mathrm{0\leq\:2\leq\:3\:(0\leq\:s\leq\:1)}$

It is clearly observed that a non-zero integer exists in such a way that if the number is divided by the integer, we get a quotient and a remainder (non-zero).

How to find HCF using Repeated Division or Euclid's Division Lemma

The major application of Euclid’s Division lemma is to determine the HCF of integers. Let’s discuss the procedure to evaluate HCF using Euclid’s Division lemma.

• Assume that we have to find the HCF of two numbers, m and l $\mathrm{(m\:>\:1)}$. Using Euclid’s Division lemma,

$$\mathrm{m\:=\:lr\:+\:s\:,\:where\:0\leq\:s\leq\:1}$$

• If $\mathrm{s\:=\:0}$, then r is the factor of m and l. If $\mathrm{s\:\neq\:0}$, then use Euclid’s Division lemma to l and s.

• We need to continue the procedure till we get the remainder as zero. The obtained divisor is the required HCF.

Solved Examples

1) Find HCF of the following numbers using Euclid’s Division lemma: 20 and 50.

Answer − As 50 > 20, hence c = 50 and d = 20.

Now, using Euclid’s Division lemma

$$\mathrm{50\:=\:20\:\times\:2\:\div\:10}$$

$$\mathrm{20\:=\:10\div\:2\:\div\:0}$$

∴ The HCF of 20 and 50 is 10.

2) Express the number 3780 as the multiplication of prime numbers.

Answer − So the prime factorization of 3780 can be expressed as

$$\mathrm{3780\:=\:2\times\:2\times\:3\times\:3\times\:3\times\:5\times\:7}$$

$$\mathrm{\Longrightarrow\:3780\:=\:2^{2}\:\times\:3^{3}\times\:5\times\:7}$$

3) Find the factors of 108.

Answer − 108 can be factorized as below.

$$\mathrm{1\times\:108\:=\:108}$$

$$\mathrm{108\times\:1\:=\:108}$$

$$\mathrm{2\times\:54\:=\:108}$$

$$\mathrm{3\times\:36\:=\:108}$$

$$\mathrm{4\times\:27\:=\:108}$$

$$\mathrm{6\times\:18\:=\:108}$$

$$\mathrm{9\times\:12\:=\:108}$$

Therefore the factors of 108 are 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, and 108.

Conclusion

The present tutorial gives a brief introduction about the prime factorization and division method to determine HCF. The basic definition and the various methods to evaluate the prime factorization have been explained in this tutorial. In addition, the repeated division or Euclid’s division theorem has been explained with proof. Moreover, some solved examples have been provided for better clarity of this concept. In conclusion, the present tutorial may be useful for understanding the basic concept of the prime factorization and division method for HCF.

FAQs

1. What is the major limitation of the prime factorization method?

The prime factorization method is time-consuming and lengthy if the given composite number is large.

2. What is the application of prime factorization?

Prime factorization is used in designing cryptography. In addition, it is utilized to evaluate HCF(Highest Common Factor) and LCM (Least Common Multiple).

3. What is the advantage of Euclid’s division lemma?

Prime factorization is a time-consuming method. However, for a large number, Euclid’s division lemma can be used to factorize the number.

4. What is the HCF of two prime numbers?

Since each prime number can only be divisible by 1 and the number itself, the HCF of two prime numbers is 1.

5. Can we determine the HCF of two negative numbers using Euclid’s division lemma?

Yes. We can evaluate the HCF of two negative numbers using Euclid’s division method.