Finding the Smallest n- digit number which is divisible exactly.

B) ii) Smallest n – digit number when divided leaves no remainder using L.C.M:

Steps to find Smallest n – digit number, which is exactly divisible by the given numbers(leaves no remainder):

1. Find out the L.C.M of the given numbers, ‘L’ .

2. Divide the smallest n– digit number with the L.C.M,

(Smallest n-digit ÷ L.C.M )  which leaves Remainder,’R’.

3. Find out the difference between L.C.M,’L’ and Remainder’R’: ( L – R ); which is to be added to the Smallest n-digit number gives the required Smallest n-digit number.

Required Smallest n-digit number = Smallest n-digit number + (L – R).

 

1) Find the least no. of 5 digits which is exactly divisible by 4, 12, 15 and 18 is?

Solution:

Given : Numbers = 4, 12, 15, 18 ; Least 5 digit number = 10000.

1. Find out the L.C.M,L of the given numbers;L.C.M( 4, 12, 15, 18 ) =  180.

2. Divide the Smallest 5 digit number,10000 with the L.C.M,L : 180 ;                           ( 10000 ÷  180 ) which leaves Remainder R = 100.

3. Subtract Remainder R =100  from L.C.M ‘L’= 180.

i.e;  L – R = 180 – 100 = 80 , which is to be added to the Smallest 5 digit number 10000 gives the required Smallest 5 digit number.

Required Smallest 5 digit number = Smallest 5 digit number + ( L – R ).

 = 10000 + 80 = 10,080, which is exactly divisible by the given numbers.                     

 

 

2) Find the smallest 3-digit number, such that they are exactly divisible by 3, 4 and 5?

Solution:

Given : Numbers = 3, 4, 5 ; Smallest 3 digit number = 100.
1.We need to find the L.C.M of the given numbers; L.C.M(3, 4, 5) = 60.

As the given numbers are Consecutive numbers, the L.C.M of the numbers is the product of the given consecutive numbers.

Therefore, L.C.M( 3, 4, 5 ) = 3 * 4 * 5 = 60 = L.

2. Divide the Smallest 3 digit number: 100, with the L.C.M,L = 60 :

 ( 100 ÷  60 ) , which leaves Remainder R = 40.

3. Find the difference between L.C.M and Remainder: L – R = 60 – 40 =20 , which is to be added to the Smallest 3 digit number, gives the Required Smallest 3 digit number.

Required Smallest 3 digit number = Smallest 3 digit number + ( L – R ).

 = 100 + 20 = 120, which is divisible exactly by the given numbers.

 

3) Find the least number of 6 digits which is exactly divisible by 30, 36, 48, 72 is?

Solution:

Given: Numbers = 30, 36, 48, 72; Smallest 6 digit number = 100000.

 

1. We need to find out L.C.M of the given numbers:

L.C.M ( 30, 36, 48, 72 ) = 720 = L.

 

 

2. Divide the Smallest 6 digit number =100000 , with the L.C.M ‘L’ = 720.

( 100000 ÷  720 ) , which leaves Remainder ‘R’ = 640.

 3.Subtract the Remainder ‘R’ = 640 from the L.C.M ’L’ = 720 .

L – R = 720 – 640 = 80 , which is to be added to the Smallest 6 digit number 100000 to get the required Smallest 6 digit number which is divisible eactly by the given numbers.

 Required Smallest 6 digit number = Smallest 6 digit number + ( L – R ).

  = 100000 + 80 = 1,00,080, which is divisible exactly by the given numbers.