Finding Greatest and Smallest n - digit number when divided leaves Remainder'K' .

C) Greatest n – digit number and Smallest n – digit number, when divided by the numbers a, b, c leaves remainder x, y, z such that a – x = b – y = c – z = K . using L.C.M method.

(i) Steps to find Greatest n – digit number when divided by the given numbers leaves remainder ’K’ respectively.

1.We need to find the L.C.M of the given numbers: L.C.M ( a, b, c ) = L.

2. Divide the greatest n – digit number with the L.C.M ‘L’:                                          (Greatest n-digit number ÷ L.C.M),which leaves Remainder ‘R’.

3. Subtract the Remainder‘R’ from the greatest n-digit number and Add Remainder 'K' to it which gives the Greatest n – digit number when divided by the numbers a, b, c, leaves remainders x, y, z  respectively.   

Required Greatest n – digit number = (Greatest n- digit number – R) + K.   

                           

 Find the greatest number of six digits which on being divided by 6, 7, 8, 9 and 10 leaves 4, 5, 6, 7 and 8 as remainder respectively.

Solution:

Given: Numbers : a, b, c, d, e = 6, 7, 8, 9 ,10 ; 

Remainders :x, y, z, u, v = 4, 5, 6, 7, 8.  Greatest 6 digit number = 9,99,999.

Check : a-x = b-y = c-z = d-u = e-v = K

              6-4 = 7-5 = 8-6 = 9-7 = 10- 8 = 2 = K.   

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

L.C.M ( 6, 7, 8, 9, 10 ) = 2520 = L.

 

2. Divide the greatest 6 digit number = 999999 with the L.C.M = 2520.                          ( 999999 ÷ 2520 ), which leaves Remainder ‘R’ = 2079.

3. Subtract the Remainders ‘R’ from the Greatest n-digit number and Add Remainder'K' to it which gives required greatest n-digit number.

Required Greatest 6 digit number = (Greatest 6 digit number – R) + K.

  = 999999 – 2079 + 2 = 9,97,922.

 

(ii) Steps to find the Smallest n –digit number when on divided by the numbers a, b, c leaves remainders x, y, z; such that a – x = b – y = c – z = ‘K’.

1. We need to find the L.C.M of the numbers; L.C.M(a, b, c) = L.

2. Divide the Smallest n – digit number with L.C.M :                                                    (Smallest n–digit number ÷ L.C.M), which leaves Remainder ‘R’.   

3. Subtract the Remainder ‘R’ from L.C.M ‘L’  : L – R .

4. Add the Remainder ‘K’ and ( L – R ) to the Smallest n – digit number which gives the required Smallest n – digit number , when on divided by the numbers a, b, c leaves remainders x, y, z respectively.

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

 

 

Find the least number of 4 digits when on divided by the numbers 12, 16, 18 and 20 leaves the remainder 21 in each case?

Solution:

Given : Numbers = 12, 16, 18, 20 ; Remainder ‘K’ = 21.

Smallest 4 digit number = 1000.

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

L.C.M ( 12, 16, 18, 20 ) = 720 = L.

2. Divide the Smallest 4 digit number = 1000 with L.C.M ’L’ = 720;                            ( 1000 ÷ 720 ) , which leaves Remainder ‘R’ = 280.

3. Subtract Remainder ‘R’ = 280 from L.C.M ‘L‘ = 720 :

 L – R = 720 – 280 = 440.

4. Add the Remainder ‘K’ and ( L – R ) to the Smallest 4 digit number = 1000  , gives the required Smallest 4 digit number, when divided by the given numbers leaves remainder 21 reapectively.

Required Smallest 4 digit number = Smallest 6 digit number + (L- R) + K.

  = 1000 + 440 + 21 = 1461.