In "Laws Of Nature" today we are going to talk about the some important arithmetic formulae on LCM and HCF, this formulae may be helpful to you, in all types of competitive exams.

# LCM AND HCF PROBLEMS TRICKS

* To find the HCF of two numbers.

Method-

Split the given numbers into prime factors and then find the product of all the prime factors, which are common to all the numbers.the product is the required HCF.

* To find the HCF of decimals.

Method-

First make the same numbers of decimals places in all the numbers,and then see the numbers as integer and find the HCF, and in the answer put the decimal places as there in each of the numbers.

* To find the LCM of two numbers.

Method-

Split the numbers into prime factors and then find the product of the highest power of all the factors that occurred in given numbers.the product will be the required LCM.

* To find the LCM of the decimals.

Method-

First make sure the same numbers of decimals places in all the numbers.then find the LCM as they are integers, and then in the answer put the decimal places as there in each of the numbers.

* HCF of the fraction=HCF of numerator/ LCM of the denominator

* LCM of the fraction= LCM of the numerator/HCF of the denominator

* Product of the numbers= HCF of the numbers× LCM of the numbers

* The greatest number that will exactly divide x,y and z is =HCF of x, y and z

* The greatest number that will divide the numbers x,y and z such that they leave remainder a,b and c respectively=

HCF of (x-a),(y-b),(z-c)

* The least number which exactly divisible by the numbers x,y and z is = LCM of x,y and z

* The least number which when divided by x, y and z leaves remainder a,b and c respectively, the special features is always observed that (x-a)=(y-b)=(z-c)=k

Then the least number is =(LCM of x y and z)-K

* The least number which when divided by the numbers x y and z leaves same remainder r in each case is =

(LCM of x y and z)+r

* The greatest number that will divide x y and z leaving the same remainder r in each case= HCF of (x-r),(y-r),(z-r)

* The greatest number that will divide x y and z leaving the same remainder in each numbers is =HCF of |(x-y)|, |(y-z)|, |(z-x)|

* To find the n digit greatest number which when divided by x y and z, leaves no remainder.

The following steps can be taken:

Step1: find LCM of x y and z that is L

Step2: divide n digit greatest number by L and find remainder R

Step3: the required number is= n digit greatest number -R

* To find the n digit smallest number which when divided by x y and z leaves no remainder.

The following steps can be taken:

Step1: find LCM of x y and z, ie. L

Step2: divide n digit smallest number by L and find the remainder R.

Step3: the required number is= n digit smallest number + (L-R)

* If there are n numbers and HCF of each pair is x and LCM of all the n numbers is y the the product of the numbers is =

[(x)^(n-1)×y]

* In the question of bell, if ringing interval are given in second, minutes and hours, then the time in which all bells rang at the same time is = ( LCM of time) + initial time

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