6 base Division Example ( Enter your problem )
  1. Addition
  2. Subtraction
  3. Multiplication
  4. Division

4. 6 base Division





1. Find division of `(123)_6` and `(4)_6`

Solution:

Solution is
  20
4123
12 =4 × 2
  3
  0=4 × 0
  3

`:.` 123 `-:` 4= 20 Remainder 3
4 table
4×1=4
4×2=12
4×3=20
4×4=24
4×5=32
4×10=40




Step by step solution

Step by step solution :
Step-1 :
Set up the problem with long division bracket. Put dividend inside bracket and divisor on outside left.
    
4123

Step-2 :
4 goes into 12 (2-times). Put a 2 in the next place of quotient and multiply 4 by 2 to get 12.
Subtract 12 from 12 to get remainder `(12-12=0)`.


  2 
4123
12 =4 × 2
  0 

Step-3 :
Now, bring down 3 from the dividend, to make 3
  2 
4123
12 =4 × 2
  3

Step-4 :
4 goes into 3 (0-times). Put a 0 in the next place of quotient and multiply 4 by 0 to get 0.
Subtract 0 from 3 to get remainder `(3-0=3)`.


  20
4123
12 =4 × 2
  3
  0=4 × 0
  3


2. Find division of `(241)_6` and `(2)_6`

Solution:

Solution is
 120
2241
2  =2 × 1
  4 
 4 =2 × 2
  1
  0=2 × 0
  1

`:.` 241 `-:` 2=120 Remainder 1
2 table
2×1=2
2×2=4
2×3=10
2×4=12
2×5=14
2×10=20




Step by step solution

Step by step solution :
Step-1 :
Set up the problem with long division bracket. Put dividend inside bracket and divisor on outside left.
    
2241

Step-2 :
2 goes into 2 (1-times). Put a 1 in the next place of quotient and multiply 2 by 1 to get 2.
Subtract 2 from 2 to get remainder `(2-2=0)`.


 1  
2241
2  =2 × 1
 0  

Step-3 :
Now, bring down 4 from the dividend, to make 4
 1  
2241
2  =2 × 1
  4 

Step-4 :
2 goes into 4 (2-times). Put a 2 in the next place of quotient and multiply 2 by 2 to get 4.
Subtract 4 from 4 to get remainder `(4-4=0)`.


 12 
2241
2  =2 × 1
  4 
 4 =2 × 2
  0 

Step-5 :
Now, bring down 1 from the dividend, to make 1
 12 
2241
2  =2 × 1
  4 
 4 =2 × 2
  1

Step-6 :
2 goes into 1 (0-times). Put a 0 in the next place of quotient and multiply 2 by 0 to get 0.
Subtract 0 from 1 to get remainder `(1-0=1)`.


 120
2241
2  =2 × 1
  4 
 4 =2 × 2
  1
  0=2 × 0
  1


3. Find division of `(31021)_6` and `(5)_6`

Solution:

Solution is
  3451
531021
23   =5 × 3
  40  
 32  =5 × 4
  42 
  41 =5 × 5
  11
  5=5 × 1
  2

`:.` 31021 `-:` 5= 3451 Remainder 2
5 table
5×1=5
5×2=14
5×3=23
5×4=32
5×5=41
5×10=50




Step by step solution

Step by step solution :
Step-1 :
Set up the problem with long division bracket. Put dividend inside bracket and divisor on outside left.
      
531021

Step-2 :
5 goes into 31 (3-times). Put a 3 in the next place of quotient and multiply 5 by 3 to get 23.
Subtract 23 from 31 to get remainder `(31-23=4)`.


  3   
531021
23   =5 × 3
  4   

Step-3 :
Now, bring down 0 from the dividend, to make 40
  3   
531021
23   =5 × 3
  40  

Step-4 :
5 goes into 40 (4-times). Put a 4 in the next place of quotient and multiply 5 by 4 to get 32.
Subtract 32 from 40 to get remainder `(40-32=4)`.


  34  
531021
23   =5 × 3
  40  
 32  =5 × 4
  4  

Step-5 :
Now, bring down 2 from the dividend, to make 42
  34  
531021
23   =5 × 3
  40  
 32  =5 × 4
  42 

Step-6 :
5 goes into 42 (5-times). Put a 5 in the next place of quotient and multiply 5 by 5 to get 41.
Subtract 41 from 42 to get remainder `(42-41=1)`.


  345 
531021
23   =5 × 3
  40  
 32  =5 × 4
  42 
  41 =5 × 5
  1 

Step-7 :
Now, bring down 1 from the dividend, to make 11
  345 
531021
23   =5 × 3
  40  
 32  =5 × 4
  42 
  41 =5 × 5
  11

Step-8 :
5 goes into 11 (1-times). Put a 1 in the next place of quotient and multiply 5 by 1 to get 5.
Subtract 5 from 11 to get remainder `(11-5=2)`.


  3451
531021
23   =5 × 3
  40  
 32  =5 × 4
  42 
  41 =5 × 5
  11
  5=5 × 1
  2





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