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

4. Hexadecimal Division





1. Find division of `(176)_16` and `(5)_16`

Solution:

Solution is
  4A
5176
14 =5 × 4
  36
 32=5 × A
  4

`:.` 176 `-:` 5= 4A Remainder 4
5 table
5×1=5
5×2=A
5×3=F
5×4=14
5×5=19
5×6=1E
5×7=23
5×8=28
5×9=2D
5×A=32
5×B=37
5×C=3C
5×D=41
5×E=46
5×F=4B
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.
    
5176

Step-2 :
5 goes into 17 (4-times). Put a 4 in the next place of quotient and multiply 5 by 4 to get 14.
Subtract 14 from 17 to get remainder `(17-14=3)`.


  4 
5176
14 =5 × 4
  3 

Step-3 :
Now, bring down 6 from the dividend, to make 36
  4 
5176
14 =5 × 4
  36

Step-4 :
5 goes into 36 (10-times). Put a 10 in the next place of quotient and multiply 5 by 10 to get 32.
Subtract 32 from 36 to get remainder `(36-32=4)`.


  4A
5176
14 =5 × 4
  36
 32=5 × A
  4


2. Find division of `(ED8)_16` and `(C)_16`

Solution:

Solution is
 13C
CED8
C  =C × 1
 2D 
24 =C × 3
  98
 90=C × C
  8

`:.` ED8 `-:` C=13C Remainder 8
C table
C×1=C
C×2=18
C×3=24
C×4=30
C×5=3C
C×6=48
C×7=54
C×8=60
C×9=6C
C×A=78
C×B=84
C×C=90
C×D=9C
C×E=A8
C×F=B4
C×10=C0




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.
    
CED8

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


 1  
CED8
C  =C × 1
 2  

Step-3 :
Now, bring down D from the dividend, to make 2D
 1  
CED8
C  =C × 1
 2D 

Step-4 :
C goes into 2D (3-times). Put a 3 in the next place of quotient and multiply C by 3 to get 24.
Subtract 24 from 2D to get remainder `(2D-24=9)`.


 13 
CED8
C  =C × 1
 2D 
24 =C × 3
  9 

Step-5 :
Now, bring down 8 from the dividend, to make 98
 13 
CED8
C  =C × 1
 2D 
24 =C × 3
  98

Step-6 :
C goes into 98 (12-times). Put a 12 in the next place of quotient and multiply C by 12 to get 90.
Subtract 90 from 98 to get remainder `(98-90=8)`.


 13C
CED8
C  =C × 1
 2D 
24 =C × 3
  98
 90=C × C
  8


3. Find division of `(4914A)_16` and `(A)_16`

Solution:

Solution is
  74ED
A4914A
46   =A × 7
  31  
 28  =A × 4
  94 
  8C =A × E
  8A
  82=A × D
  8

`:.` 4914A `-:` A= 74ED Remainder 8
A table
A×1=A
A×2=14
A×3=1E
A×4=28
A×5=32
A×6=3C
A×7=46
A×8=50
A×9=5A
A×A=64
A×B=6E
A×C=78
A×D=82
A×E=8C
A×F=96
A×10=A0




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.
      
A4914A

Step-2 :
A goes into 49 (7-times). Put a 7 in the next place of quotient and multiply A by 7 to get 46.
Subtract 46 from 49 to get remainder `(49-46=3)`.


  7   
A4914A
46   =A × 7
  3   

Step-3 :
Now, bring down 1 from the dividend, to make 31
  7   
A4914A
46   =A × 7
  31  

Step-4 :
A goes into 31 (4-times). Put a 4 in the next place of quotient and multiply A by 4 to get 28.
Subtract 28 from 31 to get remainder `(31-28=9)`.


  74  
A4914A
46   =A × 7
  31  
 28  =A × 4
  9  

Step-5 :
Now, bring down 4 from the dividend, to make 94
  74  
A4914A
46   =A × 7
  31  
 28  =A × 4
  94 

Step-6 :
A goes into 94 (14-times). Put a 14 in the next place of quotient and multiply A by 14 to get 8C.
Subtract 8C from 94 to get remainder `(94-8C=8)`.


  74E 
A4914A
46   =A × 7
  31  
 28  =A × 4
  94 
  8C =A × E
  8 

Step-7 :
Now, bring down A from the dividend, to make 8A
  74E 
A4914A
46   =A × 7
  31  
 28  =A × 4
  94 
  8C =A × E
  8A

Step-8 :
A goes into 8A (13-times). Put a 13 in the next place of quotient and multiply A by 13 to get 82.
Subtract 82 from 8A to get remainder `(8A-82=8)`.


  74ED
A4914A
46   =A × 7
  31  
 28  =A × 4
  94 
  8C =A × E
  8A
  82=A × D
  8





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