Prime Factorisation
- Prime Factorisation is the process of finding all the prime factors of a number.
- There are two methods to find the prime factors of a number-
1. Prime factorisation using a factor tree
2. Repeated Division Method
The prime factors of 36 are 2 and 3.
We can write 36 as a product of prime factors: 2 × 2 × 3 × 3
To
find prime factors using the repetitive division, it is advisable to
start with a small prime factor and continue the process with bigger
prime factors.
Highest Common Factor (HCF)
- The highest common factor (HCF) of two or more given numbers is the greatest of their common factors.
- Its other name is (GCD) Greatest Common Divisor.
Method to find HCF
Example
Find the HCF of 60 and 72.
Solution:
First, we have to find the prime factorisation of 60 and 72.
Then encircle the common factors.
HCF of 60 and 72 is 2 × 2 × 3 = 12.
Lowest Common Multiple (LCM)
The lowest common multiple of two or more given number is the smallest of their common multiples.Methods to find LCM
1. Prime Factorisation Method
To find the LCM we have to find the prime factorisation of all the given numbers and then multiply all the prime factors which have occurred a maximum number of times.Example
Find the LCM of 60 and 72.
Solution:
First, we have to find the prime factorisation of 60 and 72.
Then encircle the common factors.
To find the LCM, we will count the common factors one time and multiply them with the other remaining factors.
LCM of 60 and 72 is 2 × 2 × 2 × 3 × 3 × 5 = 360
2. Repeated Division Method
If we have to find the LCM of so many numbers then we use this method.Example
Find the LCM of 105, 216 and 314.
Solution:
Use the repeated division method on all the numbers together and divide until we get 1 in the last row.
LCM of 105,216 and 314 is 2 × 2 × 2 × 3 × 3 × 3 × 5 × 7 × 157 = 1186920
Real life problems related to HCF and LCM
Example: 1There are two containers having 240 litres and 1024 litres of petrol respectively. Calculate the maximum capacity of a container which can measure the petrol of both the containers when used an exact number of times.
Solution:
As we have to find the capacity of the container which is the exact divisor of the capacities of both the containers, i. e. maximum capacity, so we need to calculate the HCF.
Example: 2
What could be the least number which when we divide by 20, 25 and 30 leaves a remainder of 6 in every case?
Solution:
As we have to find the least number so we will calculate the LCM first.
LCM of 20, 25 and 30 is 2 × 2 × 3 × 5 × 5 = 300.
Here 300 is the least number which when divided by 20, 25 and 30 then they will leave remainder 0 in each case. But we have to find the least number which leaves remainder 6 in all cases. Hence, the required number is 6 more than 300.
The required least number = 300 + 6 = 306.