Finding the Greatest n - digit Number which is exactly divisible.

 Finding the Greatest n - digit number when divided by the numbers leaves no remainder,using L.C.M :

B) i) Greatest n-digit number, which is exactly divisible by the numbers.

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

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

2. Divide the largest n – digit number with the L.C.M( Largest n-digit ÷ L.C.M )  which leaves remainder,’R’.

3. Subtract the Remainder,’R’ from the largest n- digit number which gives the required Greatest n – digit number.

Required Largest n - digit number  = Largest n-digit number – Remainder'R' .   

 

1) Find the greatest number of 5 digits which is exactly divisible by 8, 9, 15, 21 is?

Solution:

Given : Numbers = 8, 9, 15, 21; Greatest 5 digit number = 99,999.

1.We need to find L.C.M of the given numbers: L.C.M(8, 9 15, 21 ) = 2520.

 

2. Divide the largest 5 digit number by L.C.M = 99999 ÷ 2520 , which leaves remainder ‘R’ = 1719.

3. Subtract the remainder ‘R’ from the largest 5 digit number which gives the required Greatest 5 digit number ,which is divisible exactly by the given numbers.  

99999 – 1719 = 98280, required greatest 5 digit number.

 

 

2) The greatest number of 4 digits which when divided by 2, 3, 4, 5, 6, and 7 is ?

Solution:

Given: Numbers = 2, 3, 4, 5, 6, 7 ; greatest 4 digit number = 9,999.

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

L.C.M( 2, 3, 4, 5, 6, 7 ) = 420.

2. Divide the largest 4 digit number with the L.C.M : 9999 ÷ 420 , which leaves remainder ‘R’ = 339 .

3. Subtract the obtained remainder ‘R’ from the largest 4 digit number,which gives the required Greatest 4 digit number, which is divisible exactly by the given numbers.

Required Greatest 4 digit number = Greatest 4 digit number - Remainder.

9999 – 339 = 9660, the required 4 digit greatest number.  

3) Find the greatest number of three digits which, when divided by 3, 4 and 5 leaves no remainder?

Solution:

Given: Numbers = 3, 4, 5 ; Greatest 3 digit number = 999.

1. 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 greatest n-digit number,999 with L.C.M,’L’ 60 ; 999 ÷ 60, which leaves the Remainder’R’ =  39.

3. Subtract the Remainder: 39 from the greatest 3 digit number:999 ,which gives the required Greatest 3 digit number that is divisible exactly by the given numbers.  

Required Greatest 3 digit number = Greatest 3 digit number – Remainder.

  =  999 – 39 = 960 ,which is divisible exactly by the given numbers.  

Finding the Smallest number when divided by the numbers leaves Remainder and No Remainders.

L.C.M  method to find the Smallest Numbers:

We use L.C.M method in finding the Smallest numbers in  below cases.

A) To find the Smallest or least number:

(i) The smallest number which is exactly divisible by the numbers.

(ii) The smallest number when divided by the numbers: a, b, c leaves remainder

  a) Remainder ‘K’ in each case.

  b) Remainders x, y, z respectively.

A) To find the smallest number which is exactly divisible:

(i) We need to find the L.C.M of the given numbers, which gives the least number of the given numbers.

1) Find the smallest number which is exactly divisible by 72, 90 and 120.
Solution:

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

L.C.M( 72, 90, 120) = 360.

Therefore, L.C.M is the smallest number which is divisible exactly by the given numbers = 360.

 

2) Find the smallest number which is divisible exactly by the numbers 7, 18, 56, 36 is?

Solution:

We need to find the L.C.M of the given numbers: L.C.M ( 7, 18, 56, 36 ) =

 

Therefore, L.C.M is the smallest number which is divisible exactly by the numbers 7, 18, 56, 36 = 504.

A) (ii) (a) Steps to find the smallest number when divided by the numbers leaves remainder ‘K’ in each case:

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

2. The required Smallest number is obtained by adding the L.C.M obtained in the step-1 and the Remainder:‘K’ given.

Smallest number = L + K.

 

3) Find the least number when divided by 36, 24, 16 leaves 11 as remainder in each case?

Solution:

Given : Remainder ‘K’ = 11 ;  Numbers = 36, 24, 16.

(i) We need to find the L.C.M : L of the given numbers.

L.C.M(36, 24 ,16) =  144.

The least number = L.C.M of numbers’L’ + Remainder’K’.

                                = 144 + 11 =  155.

 

4) What is the least number, which when divided by 98 and 105 has in each case 10 as remainder?

Solution:

Given : Numbers = 98 and 105;  Remainder ’K’ = 10.

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

    L.C.M(98, 105) ‘L’= 1470.

Therefore, Smallest number = L.C.M ‘L’ + Remainder ‘K’

Smallest Number = 1470 + 10 = 1480.

 

5) Find the least number which being divided by 2, 3, 4, 5, 6, leaves in each case a remainder 1, but when divided by 7 leaves no remainder.

Solution :

Given : Numbers = 2, 3, 4, 5, 6 ; Reaminder ‘K’ = 1.

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

L.C.M( 2, 3, 4, 5, 6) ,’L’ = 60.

Therefore, the smallest number = L + K = 60 + 1 = 61.

Here, we have to check the second condition; the smallest number when divided by 7 leaves no remainder.

The smallest number we got is 61, when divided by 7 leaves remainder; which doesn’t satisfy the given condition. So, we have to find the multiples of Smallest number.

Multiples of 60 = 60, 120, 180, 240, 300, 360, 420, 480, 540, 600.

We have to check : The multiples of 60 + Remainder ‘K’ is to be divisible by 7 exactly.

Therefore, the smallest number is 301, which is divisible by 7 exactly.

 

(A)(ii)(b) Steps to find the Smallest number when divided by the numbers a, b, c leaves remainders x, y, z respectively:

1.Find out the L.C.M of the given numbers; L.C.M(a, b c) = L.

2. Check the difference between the numbers and the remainders, which is equal to ‘K’.

3. The difference between the L.C.M ‘L’ and the Remainder ’K’ gives the  required Smallest number.  

Smallest number = L – K .

 

6) What is the smallest number when divided by the numbers 20, 48, 36 leaves remainders 13, 41, 29 respectivey.

Solution:

Given : Numbers: a, b, c = 20, 48, 36 ; Remainders: x, y, z = 13, 41, 29.

1. L.C.M( 20, 48, 36 ) = 720.

2. The difference between the numbers and the remainders = a-x = b-y = c-z =  K, say.

i.e; 20 – 13 = 48 – 41 = 36 – 29 = 7 = K.

3. The Smallest number = L – K = 720 – 7 = 713.

 

8) Find the smallest number which on dividing by 6 8 12 leaves the remainder 4 6 10 respectively?

Solution:

Given: Numbers: a, b, c = 6, 8, 12 ; Remainders: x, y, z = 4, 6, 10.

1. L.C.M( a, b, c ) = L.C.M(6, 8, 12) = 24.

2. The difference between the numbers and remainders: a – x = b – y = c – z = K ; 6 – 4 = 8 – 6 = 12 – 10 = 2 = K.

3. The Smallest number = L – K = 24 – 2 = 20.

Relation between H.C.F and L.C.M

Relation Betweeen H.C.F and L.C.M:

Case(i) Product of two numbers = Product of their H.C.F. and L.C.M.

Case(ii) There are n numbers. If the HCF of each pair is x and the LCM of all the n numbers is y, then the product of n numbers is given by    or Product of ‘n’ numbers = (HCF of each pair)^n-1 × (LCM of n numbers).

 

Find the numbers:

1) The LCM of two numbers is 64699, their GCM is 97 and one of the numbers is 2231. Find the other.

Solution:

Product of two numbers = Product of their H.C.F and L.C.M

Given Data:  L.C.M = 64699 ; H.C.F = 97.

 Let the number be : A = 2231.

 Other number be : B =?

We know, A× B = H.C.F × L.C.M

            2231 × B = 97 × 64699

           B =  (97 × 64699) / 2231 = 2813.

Therefore,the other number B = 2813.

 

2) The L.C.M of two numbers is 2079, and their H.C.F is 27.If one of the number is 189 then find other number?

Solution:

Given Data : L.C.M = 2079 ; H.C.F = 27.

Let one number be A = 189;

The other number be B =?

We know, A× B = H.C.F × L.C.M

                189 × B = 27 × 2079

       B = (27 × 2079) / 189 = 297.

Therefore, the other number B = 297.         

 

3) The ratio of the two numbers is 5:6. The L.C.M of two numbers is 480.Find out their H.C.F?

Solution:

Given: The ratio of the two numbers = 5:6 ; L.C.M = 480.

Assume the number be ‘x’, then the two numbers are = 5x and 6x.

We need to find out H.C.F of two numbers : H.C.F(5x ,6x) = x.

Factors of 5x = x ,5 ,

Factors of 6x = x, 6.

Therefore, H.C.F( 5x,6x) = x.

We know, Product of two numbers = Product of their H.C.F and L.C.M.

So,   (5x) × (6x) = x × 480

         x = 16 .

Therefore, H.C.F = 16.

  

4) There are 4 numbers. The HCF of each pair is 3 and the LCM of all the 4 numbers is 116. What is the product of 4 numbers?
Solution:

Given : H.C.F of each pair =3 ; L.C.M of 4 numbers = 116 ; n = 4.

We Know;

Product of ‘n’ numbers = (HCF of each pair)^n-1 × (LCM of n numbers).

Product of 4 numbers = (H.C.F of 2 pairs)^4-1 × ( L.C.M of 4 numbers)

Product of 4 numbers = ( 3 ) ³ ×  116

                                       =  27 × 116 = 3132.
Therefore, Product of 4 numbers = 3132.

H.C.F of Decimal Numbers.

H.C.F of Decimal Numbers:

Convert the given numbers into same number of decimal places in all the given numbers; then find their H.C.F as if they were integers, and mark in the result as many decimal places as there are in each of the numbers.

Find the H.C.F of Decimal Numbers:

1) 16.5, 0.45 and 15.
Answer:
The given numbers are equivalent to 16.50, 0.45 and 15.00
Step I:   First we find the H.C.F of 1650, 45 and 1500.

 

Step II:  The required H.C.F = 0.15.

                 (or)

(i) Convert the decimal numbers into fractions.

16.5 = 165/10 ; 0.45 = 45/100 ; 15 = 15/1

(ii) H.C.F ( x/a , y/b, z/c ) = H.C.F (x,y,z) / L.C.M ( a,b,c ).

Therefore, H.C.F ( 165/10, 45/100, 15/1 )

= H.C.F (165, 45, 15) / L.C.M ( 10, 100, 1 ).

(iii) H.C.F (165, 45, 15) = 15

       L.C.M ( 10, 100, 1 ) = 100.

(iv) H.C.F ( 165/10, 45/100, 15/1 )

    =  15/ 100 = 0.15 .

 

2) 1.7, 0.51, and 0.153.

Answer:
Step (i): Converting given decimal numbers into like decimals:1700, 510, 153.

 

Step(ii) We find the HCF of 1700, 510 and 153.

Factors of 1700 = 1, 17, 100, 1700.

Factors of 510 = 1, 3, 17, 30, 170, 510.

Factors of 153 = 1, 9, 17, 153.

H.C.F( 1700, 510, 153 ) = 17.

 

Step (iii): The required H.C.F = 0.017.

                   (or)

(i) Converting given Decimal numbers into fractions.

   1.7 = 17/10 ; 0.51= 51/100; 0.153= 153/1000.

 (ii) H.C.F ( x/a , y/b, z/c ) = H.C.F (x,y,z) / L.C.M ( a,b,c ).

Therefore, H.C.F ( 17/10, 51/100, 153/1000 )

= H.C.F (17, 51, 153) / L.C.M ( 10, 100, 1000 ).

(iii) H.C.F (17, 51, 153) = 17

       L.C.M ( 10, 100, 1000 ) = 1000.

(iv) H.C.F ( 17/10, 51/100, 153/1000 )

      H.C.F = 17 /1000 = 0.017.