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.
Least Common Multiple(LCM):
The least number which is exactly divisible by each one of given numbers is called LCM of these numbers.
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
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
LCM =(LCM of 2,8,16,10)/(HCF of 3,9,81,27)=80/3
1.Find the largest number of four digits which is exactly divided by 15,18 and 27.
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
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?
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?
Required Number: LCM of (15,30 and 60) + 8
Hence, Number is 68.