Prime Factorization of HCF and LCM

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Prime factorization method and the Division Method are the two main techniques that are used to determine the HCF (Highest Common Factor) and the LCM (Least Common Multiple) of the numbers. The largest number that can divide each of the given numbers without leaving any remainder is known as the highest common factor (HCF) of two or more numbers, For example, the HCF of 4, 8, and 6 is 2. The lowest of the common multiples of two or more numbers is called the least common multiple (LCM), For example, the LCM of 10, 20, and 15 is 60. Both the methods are shown then with numerous instances.

Table of Content

How to find HCF and LCM?
Finding LCM by Prime Factorisation Method
Finding LCM by Division Method.
Finding HCF by Prime Factorisation Method.
Finding HCF by Division Method.
Things to Remember
Sample Questions

Key Terms: HCF, LCM, Prime, Factorisation, Division, Number, Methods

How to find HCF and LCM?

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We can obtain the LCM and HCF of the mentioned natural numbers by two methods i.e., by division method or by prime factorization method. In the prime factorization method, the mentioned numbers are shown as the multiple prime factors. While in the division method, the mentioned numbers are divided by the least common factor and the process is repeated until the remainder is zero. Also, Prime numbers are the kind of numbers that have just two factors i.e. the number itself and one.

Finding LCM by Prime Factorisation Method

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In this method, the mentioned natural numbers are shown as a multiple of prime factors. The lowest common multiple shall be the multiple of all prime factors with the highest degree of power.

Example: Find the LCM of 20 and 12 by the prime factorisation method.

Solution: We can find the LCM of 20 and 12 by:

Step 1: For finding the LCM of 20 and 12, we write each number as a multiple of prime factors.

20 = 2 × 2 × 5 = 2² × 5

12 = 2 × 2 × 3 = 2² × 3

Step 2: We now Multiply all the prime factors with their highest degree.

Here we have got 2 that has the highest power 2 and other prime factors 5 and 3. We shall Multiply all these to get LCM.

LCM of 20 and 12 = 2 × 2 ×3 × 5

= 2² × 3 × 5

= 60

Finding LCM by Division Method

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In the division method, we divide the mentioned numbers by a common prime number until the remainder is found to be a prime number or one. LCM is the product obtained by multiplying all divisors and the remaining prime numbers.

Example: Find the LCM of 24 and 15 by the division method.

Solution: The LCM of these numbers can be found by:

Step 1: First, we shall divide the mentioned numbers by the least prime number.

As 2 is the least number here, it will divide 24.

Division Method

Step 2: Now, we shall write the quotient and the number that can’t be divided by the above prime number in the second row.

In the second row, we shall write the quotient we get after dividing 24 by 2. Since 15 is not divisible by 2, we shall write 15 in the second row as it is.

Step 3: We shall divide the numbers with another smallest prime number.

LCM

Step 4: Continue dividing it until the remainder is found to be a prime number or 1.

Prime Factorization Method

Step 5: We shall now Multiply all the remaining prime numbers (if any) and the divisors to obtain the LCM.

LCM of 24 and 15= 2 × 2 × 2 × 3 × 5 = 2³ × 3 × 5 = 120

Finding HCF by Prime Factorisation Method

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The mentioned natural numbers shall be written as the multiple of prime factors. We shall multiply all the common prime factors with the lowest degree of power to get the highest common factor.

Example: Find the HCF of 12 and 20 by the prime factorisation method.

Solution: The steps for the same are:

Step 1: To find HCF of 12 and 20, we shall write each number as a multiple of prime factors.

20 = 2 × 2 × 5 = 2² × 5

12 = 2 × 2 × 3 = 2² × 3

Step 2: Now we shall Multiply all the common prime factors with the lowest degree of power.

Here we have got only 2 as a common prime factor with a power of 2.

HCF of 12 and 20 = 2² = 4

Finding HCF by Division Method

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In the division method for finding the HCF, we shall divide the largest number by the smallest number among the mentioned numbers until we find the remainder to be zero. The last divisor found will become the HCF of the mentioned numbers.

Example: Find the LCM of 15 and 24 by the division method.

Solution: The steps for finding LCM of 15 and 24 by the division method are as follows:

Step 1: Divide the largest number by the smallest number.

Here, the smaller number is 15 and the largest number is 24. We shall Divide 24 by 15

Prime factorization solution step 1

Step 2: Take the divisor as the new dividend and the remainder as the new divisor, which means, to divide the first divisor by the first remainder.

Prime Factorization

Step 3: Continue this until the remainder is found to be zero and the last divisor found will be the HCF of the mentioned numbers.

HCF

Therefore, the HCF of 15 and 24 is 3.

We can also divide both the numbers by the smallest common prime factor until there are no more common prime factors. Multiply all divisors to get the HCF of mentioned numbers.

Example: Find the HCF of 15 and 24.

Step 1: We shall Divide the mentioned numbers by the smallest common prime factor.

Here, 3 is the smallest common prime factor of 15 and 24.

factor

Step 2: We shall continue until there are no more common prime factors to be found. Then we shall multiply all the divisors.

Prime Factorization

The Division of 15 and 24 by 3 will leave 5 and 8 as their respective remainders. 8 and 5 do not have a common prime factor.

So, the HCF of 15 and 24 is 3.

Things to Remember

The Prime factorization method and the Division Method are the two main techniques that are used to determine the HCF ( Highest Common Factor) and the LCM (Least Common Multiple) of the numbers.
The largest number that can divide each of the given numbers without leaving any remainder is known as the highest common factor (HCF) of two or more numbers,
The lowest of the common multiples of two or more numbers is called the least common multiple (LCM),
In the prime factorization method, the mentioned numbers are shown as the multiple prime factors.
In the division method, the mentioned numbers are divided by the least common factor and the process is repeated until the remainder is zero.

Sample Questions

Ques: Find the LCM of 60, 80 and 108 by listing multiples. (5 marks)

Ans: We shall find the LCM of 60, 80 and 108 by the following steps.

Step 1: We shall List the multiples of 60, 84, and 108:

Multiples of 60: 60, 120, 180, 240, 300, 360, 420, 480, 540, ….

Multiples of 84: 84, 168, 252, 336, 420, 504, 588, 672, …

Multiples of 108: 108, 216, 324, 432, 540, 648, 756, 864, 972, ….

Step 2: We shall Find the common multiples from the multiples of 60, 84, and 108:

The common multiples are 3780, 7560, …

Step 3: We shall Determine the least common multiple from the multiples of 60, 84, and 108:

Thus, the least common multiple of 60, 84, and 108 is 3780.

So the LCM of 60, 84, and 108 is 3780.

Ques: Find the LCM of 60, 84 and 108 by Prime Factorisation Method. (5 marks)

Ans: To find the LCM of 60, 84, and 108 using the prime factorization method, firstly, we write the prime factorization of 60, 84, and 108.

The prime factorization of 60 is 22 × 31 × 51

The prime factorization of 84 is 22 × 31 × 71

The prime factorization of 108 is 22 × 33

Therefore, the LCM of 60, 84, and 108 is found by multiplying the prime factors raised to their highest power.

So the least common multiple of 60, 84 and 108 = 22 × 33 × 51 × 71 = 3780.

Hence, LCM of 60, 84, and 108 is 3780.

Ques: Find the LCM of 60, 84 and 108 by Division Method. (5 marks)

Ans: To find out the LCM of 84, 108 and 60 by using the division method, we shall divide the numbers 84, 108 and 60 by the prime factors. So, the product of the divisors provides the result of the LCM of 84, 108 and 60.

Lcm

Therefore, LCM of 60, 84 and 108 = 2 × 2 × 3 × 3 × 3 × 5 × 7 = 3780.

Ques: What is the division method for finding the LCM and HCF? (3 marks)

Ans: In the division method for finding the LCM and HCF, there are four ways that involve mathematical operations which are divide, multiply, subtract, and bring down. The first step involves the division of the given number by the divisor by the identification of the capable numbers. additionally, in the following step, the divisor and number are multiplied to get the number to be deducted from the dividend. The subtracted number is brought down as the remainder and another number from the dividend and this operation is continued till a 0 remainder is received or a number that's lower than the divisor.

Ques: What are some of the properties of HCF? (3 marks)

Ans: The properties of HCF of two or more numbers are mentioned below:

In HCF, each of the numbers is divided without any remainder.
HCF is a factor of each of the given numbers.
It is always found to be either less than or equal to one but it can never be greater.
If the numbers mentioned are prime numbers, then the HCF is always equal to one.

Ques: What are some of the properties of LCM? (3 marks)

Ans: The properties of LCM of two or more mentioned numbers are given below:

LCM can not be less than any of the given numbers. For example, the LCM of 4 and 7 is equal to 28 and that is not smaller than any of the given numbers.
If a case arises in which one number is a factor of another in a pair of given numbers, then The LCM will be is the greater number between the two of them. For example, the LCM of 4 and 2 will be 4 which is the greater number itself.

Ques: What are some of the differences between a factor and a multiple? (3 marks)

Ans: The number which can be divided for a given number of times by another number and leave no remainder is called the multiple of that number while the factor is the number which after being divided by a number leaves no remainder behind. For a given number, the factors can be finite whereas the multiples are infinite. Another crucial difference is that the resulting factor is always lower than or equal to the mentioned number while the resulting multiples are higher than or equal to the mentioned number.

Ques: Find the HCF and LCM of 180 and 24 using prime factorisation. (3 marks)

Ans: HCF of 180 and 24:

The prime factors of 180= 2 × 2 × 3 × 3 × 5

The prime factors of 24= 2 × 2 × 2 × 3

The HCF is the product of the common prime factors of the given numbers.

Therefore, the HCF of 180 and 24 is 2 × 2 × 3 = 12

Ques: Find the LCM of 180 and 24 using the prime factorisation. (3 marks)

Ans: LCM of 180 and 24:

The prime factors of 180 = 2 × 2 × 3 × 3 × 5

The prime factors of 24 = 2 × 2 × 2 × 3

Taking all the prime factors from both the numbers only once, we have: Common prime factors (2 × 2 × 3) × Uncommon prime factors (2 × 3 × 5) = 360

Therefore, the LCM of 180 and 24 = 360.

Ques: Find the product of two numbers whose LCM and HCF are 12 and 2 respectively. (3 marks)

Ans: We know that the HCF and LCM formula for two numbers is HCF × LCM = product of the numbers.

Therefore, the product of numbers = 12 × 2 = 24

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Prime Factorization of HCF and LCM

How to find HCF and LCM?

Finding LCM by Prime Factorisation Method

Finding LCM by Division Method

Finding HCF by Prime Factorisation Method

Finding HCF by Division Method

Things to Remember

Sample Questions

CBSE X Related Questions

1.
Two water taps together can fill a tank in $8\frac{8}{9}$ hours. The tap of larger diameter takes 4 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

2.
PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If OP = 13 cm, then find the length AB and PA.

3.
The dimensions of a window are $156\text{ cm} \times 216\text{ cm}$. Arjun wants to put grill on the window creating complete squares of maximum size. Determine the side length of the square and hence find the number of squares formed.

4.
In the given figure, $AB \parallel DE$ and $AC \parallel DF$. Show that $\Delta ABC \sim \Delta DEF$. If $BC = 10\text{ cm}$, $EB = CF = 5\text{ cm}$ and $AB = 7\text{ cm}$, then find the length $DE$.

5.
If the zeroes of a polynomial p(x) are $-3$ and 8, then p(x) equals

6.
A chord of a circle, of radius 14 cm, subtends an angle of $60^\circ$ at the centre. Find the area of the smaller sector and perimeter of the smaller segment.

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Prime Factorization of HCF and LCM

How to find HCF and LCM?

Finding LCM by Prime Factorisation Method

Finding LCM by Division Method

Finding HCF by Prime Factorisation Method

Finding HCF by Division Method

Things to Remember

Sample Questions

CBSE X Related Questions

1.Two water taps together can fill a tank in $8\frac{8}{9}$ hours. The tap of larger diameter takes 4 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

2.PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If OP = 13 cm, then find the length AB and PA.

3.The dimensions of a window are $156\text{ cm} \times 216\text{ cm}$. Arjun wants to put grill on the window creating complete squares of maximum size. Determine the side length of the square and hence find the number of squares formed.

4.In the given figure, $AB \parallel DE$ and $AC \parallel DF$. Show that $\Delta ABC \sim \Delta DEF$. If $BC = 10\text{ cm}$, $EB = CF = 5\text{ cm}$ and $AB = 7\text{ cm}$, then find the length $DE$.

5.If the zeroes of a polynomial p(x) are $-3$ and 8, then p(x) equals

6.A chord of a circle, of radius 14 cm, subtends an angle of $60^\circ$ at the centre. Find the area of the smaller sector and perimeter of the smaller segment.

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Two water taps together can fill a tank in $8\frac{8}{9}$ hours. The tap of larger diameter takes 4 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

2.
PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If OP = 13 cm, then find the length AB and PA.

3.
The dimensions of a window are $156\text{ cm} \times 216\text{ cm}$. Arjun wants to put grill on the window creating complete squares of maximum size. Determine the side length of the square and hence find the number of squares formed.

4.
In the given figure, $AB \parallel DE$ and $AC \parallel DF$. Show that $\Delta ABC \sim \Delta DEF$. If $BC = 10\text{ cm}$, $EB = CF = 5\text{ cm}$ and $AB = 7\text{ cm}$, then find the length $DE$.

5.
If the zeroes of a polynomial p(x) are $-3$ and 8, then p(x) equals

6.
A chord of a circle, of radius 14 cm, subtends an angle of $60^\circ$ at the centre. Find the area of the smaller sector and perimeter of the smaller segment.