**Highest Common Factor (HCF) :**

The HCF** **of two or more numbers is the greatest number which divides each number exactly.

How to find H.C.F of two numbers:

- Factor Method: Make the factors of numbers as product of prime numbers and the product of least power of common prime numbers will Be HCF.
- Division Method: If we have two numbers , divide larger one with smaller number ,then divide the divisor by remainder and so on till we get “zero” as remainder.The last divisor is HCF of numbers.

**Example:**

**Least Common Multiple(LCM):**

The least number which is exactly divisible by each one of given numbers is called LCM of these numbers.

**Example:**

**Some Important Formulae for HCF and LCM:**

**N1 x N2 = LCM x HCF**

Where N1 and N2 are numbers.

**HCF Of Fractions :**

HCF=(HCF of Numerators)/(LCM of Denominators)

**Example:** Find HCF of 2/3,8/9,16/81 and 10/27

**Solution:**

HCF =(HCF of 2,8,16,10)/(LCM of 3,9,81,27)=**2/81**

**LCM Of Fractions :**

LCM=(LCM of Numerators)/(HCF of Denominators)

**Example:** Find LCM of 2/3,8/9,16/81 and 10/27

**Solution:**

LCM =(LCM of 2,8,16,10)/(HCF of 3,9,81,27)=**80/3**

** **

**Exercises**

**1.Find the largest number of four digits which is exactly divided by 15,18 and 27.**

**Ans:**

Largest number of four digits: 9999

Required number must be divisible by LCM of 15,18 and 27,then LCM of(15,18 and 27) =270

Dividing 9999 by 270, we get 9 remainder

Hence, Required number: (9999-9)=**9990**

**2.Find the smallest number of four digits which is exactly divided by 8,16,and 24**

**Ans:**

Smallest number of four digits :1000

Required number must be divisible by LCM of 8,16 and 24 then LCM of (8,16,24)=48

Dividing 1000 by 48 , we get 40 remainder

Hence, Required number= 1000+(48-40) =**1008**

**3.Traffic lights changes after every 25 sec, 50 sec and 75 sec respectively .If they all change at 01:20:00 Hours, then at what time these will change again at same time?**

**Ans:**

Changing interval of light at same time : LCM of (25,50 and 75)=150 sec

So, light will again blink at same time after 150 sec

- 2 min and 30 sec

Hence, light will change again simultaneously at **01:22:30** hours.

**4.What is the least number which divided by 15,30,and 60 leaves 8 as remainder in each case?**

**Ans:**

Required Number: LCM of (15,30 and 60) + 8

- 60+8
- 68

Hence, Number is **68.**