HCF and LCM

HCF

HCF is the acronym of Highest Common Factor. It is calculated for two or more numbers. Highest Common Factor (HCF) of two or more numbers is the greatest number that divides the given numbers exactly

In class 4, we have learnt to find HCF of two numbers upto 2-digit by prime factorization method. Some examples for revision have been given below:

HCF by Prime Factorization Method

Let us see the given examples:

Example 1

Find HCF of 8 and 12 by prime factorization method.

Solution

Thus, 4 is the HCF of 8 and 12.

Example 2

Find HCF of 24 and 40 by prime factorization method.

Solution

Thus, 8 is the HCF of 24 and 40.

Finding HCF of three numbers upto 2-digit

• By prime factorization method

Example

Find HCF of 16, 24 and 48 by prime factorization method

Solution

Thus, 8 is the HCF of 16, 24 and 48.

HCF and LCM

LCM

LCM is the acronym of Least Common Multiple. It is calculated for two or more numbers. Least Common Multiple (LCM) of two or more numbers is the smallest number among the common multiples.

To find LCM, follow these steps:

 1. Find multiples of given numbers.

2. Choose smallest common number among the calculated multiples.

 Chosen smallest common number is the required LCM.

Example 1

Find LCM of 8, 12 and 24

Solution

Smallest common multiple of 8, 12 and 24 is 24.

So, LCM of 8, 12 and 24 is 24.

Finding LCM of four numbers upto 2-digit

LCM by prime factorization method

To find LCM by prime factorization method, follow these steps:

1. Find prime factors of all the given numbers.

2. Find product of common prime factors in two or more numbers and non-common prime factors.

3. Multiply all common and non-common prime factors.

Example 1

Find LCM of 8, 12 and 24 by prime factorization method.

Solution

Prime factorization of 8 = 2 × 2 × 2

Prime factorization of 12 =      2 × 2 × 3

Prime factorization of 24 = 2 × 2 ×2 × 3

Product of common prime factors = 2 × 2 × 2 × 3= 24

Hence, LCM of 8, 12 and 24 is 24

Example 2

Find LCM of 21, 27, 51 and 81 by prime factorization method

Solution

Prime factorization of 21 =            3 × 7

Prime factorization of 27= 3 × 3 × 3

Prime factorization of 51 =        ×  3 × 17

Prime factorization of 81 = 3 × 3 × 3 × 3

Product of common prime factors = 3 × 3 × 3 = 27

Product of non-common prime factors = 7 × 17 × 3 = 357

Product of common and non-common prime factors

= 27 × 357 = 9639

Hence, LCM of 21, 27, 51 and 81 is 9639

HCF and LCM

Solve real life problems involving HCF and LCM

Example 1 (HCF)

Find the maximum length of a measuring tape that can exactly measure 18, 24 and 30 metre of wires?

Solution

We have to find HCF of 18, 24 and 30 to calculate the exact length of measuring tape.

Prime factorization of 18 = 2 × 3 × 3

Prime factorization of 24 = 2 × 2 × 2 × 3

Prime factorization of 30 = 2 × 3 × 5

Common factors of 18, 24 and 30 = 2 , 3

Product of common factors = 2 × 3=6

Thus, 6 metres long measuring tape is required to measure 18, 24 and 30 metre of wires exactly.

Example 2 (LCM)

How much minimum distance can exactly be measured with 10, 20, 25 and 30 metre long strings?

Solution

We have to find LCM to calculate the required distance:

 LCM = 2 × 5 × 2 × 3 × 5 = 300

 So, required distance is 300 metres

HCF and LCM

Solve real life problems involving HCF and LCM

Example 1 (HCF)

Find the maximum length of a measuring tape that can exactly measure 18, 24 and 30 metre of wires?

Solution

We have to find HCF of 18, 24 and 30 to calculate the exact length of measuring tape.

Prime factorization of 18 = 2 × 3 × 3

Prime factorization of 24 = 2 × 2 × 2 × 3

Prime factorization of 30 = 2 × 3 × 5

Common factors of 18, 24 and 30 = 2 , 3

Product of common factors = 2 × 3=6

Thus, 6 metres long measuring tape is required to measure 18, 24 and 30 metre of wires exactly.

Example 2 (LCM)

How much minimum distance can exactly be measured with 10, 20, 25 and 30 metre long strings?

Solution

We have to find LCM to calculate the required distance:

 LCM = 2 × 5 × 2 × 3 × 5 = 300

 So, required distance is 300 metres

HCF and LCM

Solve real life problems involving HCF and LCM

Example 1 (HCF)

Find the maximum length of a measuring tape that can exactly measure 18, 24 and 30 metre of wires?

Solution

We have to find HCF of 18, 24 and 30 to calculate the exact length of measuring tape.

Prime factorization of 18 = 2 × 3 × 3

Prime factorization of 24 = 2 × 2 × 2 × 3

Prime factorization of 30 = 2 × 3 × 5

Common factors of 18, 24 and 30 = 2 , 3

Product of common factors = 2 × 3=6

Thus, 6 metres long measuring tape is required to measure 18, 24 and 30 metre of wires exactly.

Example 2 (LCM)

How much minimum distance can exactly be measured with 10, 20, 25 and 30 metre long strings?

Solution

We have to find LCM to calculate the required distance:

 LCM = 2 × 5 × 2 × 3 × 5 = 300

 So, required distance is 300 metres

HCF and LCM

Solve real life problems involving HCF and LCM

Example 1 (HCF)

Find the maximum length of a measuring tape that can exactly measure 18, 24 and 30 metre of wires?

Solution

We have to find HCF of 18, 24 and 30 to calculate the exact length of measuring tape.

Prime factorization of 18 = 2 × 3 × 3

Prime factorization of 24 = 2 × 2 × 2 × 3

Prime factorization of 30 = 2 × 3 × 5

Common factors of 18, 24 and 30 = 2 , 3

Product of common factors = 2 × 3=6

Thus, 6 metres long measuring tape is required to measure 18, 24 and 30 metre of wires exactly.

Example 2 (LCM)

How much minimum distance can exactly be measured with 10, 20, 25 and 30 metre long strings?

Solution

We have to find LCM to calculate the required distance:

 LCM = 2 × 5 × 2 × 3 × 5 = 300

 So, required distance is 300 metres

HCF and LCM

Solve real life problems involving HCF and LCM

Example 1 (HCF)

Find the maximum length of a measuring tape that can exactly measure 18, 24 and 30 metre of wires?

Solution

We have to find HCF of 18, 24 and 30 to calculate the exact length of measuring tape.

Prime factorization of 18 = 2 × 3 × 3

Prime factorization of 24 = 2 × 2 × 2 × 3

Prime factorization of 30 = 2 × 3 × 5

Common factors of 18, 24 and 30 = 2 , 3

Product of common factors = 2 × 3=6

Thus, 6 metres long measuring tape is required to measure 18, 24 and 30 metre of wires exactly.

Example 2 (LCM)

How much minimum distance can exactly be measured with 10, 20, 25 and 30 metre long strings?

Solution

We have to find LCM to calculate the required distance:

 LCM = 2 × 5 × 2 × 3 × 5 = 300

 So, required distance is 300 metres