The basic arithmetic operations such as Addition, Subtraction, Multiplication and Division for hexadecimal numbers are performed here. In hexadecimal number system there are 16 digits starting with 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E and F. Here the addition of hexadecimal numbers is performed first.

Hexadecimal Addition

Hexadecimal Addition is achieved following by the given two steps given below.

1. Add the right most digits of each hexadecimal numbers.
2. Find the modulo of the sum of digits means divide the sum by 16 and the remainder so obtained is hexadecimal equivalent of the sum and the quotient is added to the next sum of digits from right side. Perform the second step until we add the last left most digits. Let us solve some examples to understand the processes.

Question 1: Find the sum of the hexadecimal numbers A23₁₆ and 1B1₁₆?

Solution: The addition of A23₁₆ and 1B1₁₆ is shown below.

Digit (Starting from Right)	Remainder	Quotient
3 + 1 = 4	4 % 16 = 4	4 / 16 = 0
(Quotient added) 2 + B + 0 = 13	13 % 16 = 13 (D)	13 / 16 = 0
(Quotient added) A + 1+ 0 = 11	11 % 16 = 11(B)	11 /1 6 = 0

Thus the addition of A23₁₆ + 1B1₁₆ = BD4₁₆.

Question 2: Find the sum of the hexadecimal numbers 2AC₁₆ and FAB₁₆?

Solution: The solution of the above problem is shown below:

Digit (Starting from Right)	Remainder	Quotient
C + B = 23	23 % 16 = 7	23 / 16 = 1
(Quotient added) A + A + 1= 21	21 % 16 = 5	21 / 16 = 1
(Quotient added) 2 + F + 1= 18	18 % 16 = 2	18 / 16 = 1
(Quotient added) 0 + 0 + 1 = 1	1 % 16 = 1	1 / 16 = 0

Thus the sum of the hexadecimal numbers 2AC₁₆ and FAB₁₆ is 1257₁₆.

Question 3: Find the addition of A2B.D4₁₆ and 247.65₁₆?

Solution: The solution of the above problem is performed as:

Digit (Starting from Right)	Remainder	Quotient
4 + 5 = 9	9 % 16 = 9	9 / 16 = 0
(Quotient added) D + 6 + 0 = 19	19 % 16 = 3	19 / 16 = 1
(Quotient added) B + 7 + 1 = 19	19 % 16 = 3	19 / 16 = 1
(Quotient added) 2 + 4 + 1 = 7	7 % 16 = 7	7 / 16 = 0
(Quotient added) A + 2 + 0 = 12	12 % 16 = 12(C)	12 / 16 = 0

Thus the sum is C73.39₁₆.

Hexadecimal Subtraction

Subtraction of hexadecimal numbers is obtained by following the given steps.

1. First find 15’s complement or 16’s of the subtrahend.
2. Add minuend and 15’s complement or 16’s of the subtrahend using the above addition steps of hexadecimal numbers.
3. Discard the left most end carry and add 1 to the right most digit (only in case of 15’s complement) and that will be the final solution.

Question 4: Subtract the hexadecimal numbers ABC₁₆ and A3B₁₆?

Solution: here we will find the subtraction of the two given numbers using 15’s complement. So the solution for 15’s complement of A3B₁₆ is as follows:

15 – A = 5

15 – 3 = C (in hexadecimal number system 12 = C)

15 – B = 4

So the 15’s complement of A3B₁₆ is 5C4₁₆.

Now add ABC₁₆ and 5C4₁₆.

Digit (Starting from Right)	Remainder	Quotient
C + 4 = 16	16 % 16 = 0	16 / 16 = 1
(Quotient added) B + C + 1 = 24	24 % 16 = 8	24 / 16 = 1
(Quotient added) A + 5 + 1 = 16	16 % 16 = 0	16 / 16 = 1
(Quotient added) 0 + 0 + 1 = 1	1 % 16 = 1	1 / 16 = 0

The remainder is 1080₁₆. Now we have to discard the left most end carry 1 and then we have to add 1 to the right most digit.

Thus 1080₁₆ becomes 80₁₆ + 1 = 81₁₆

Hence, ABC₁₆ – A3B₁₆ = 81₁₆.

We can solve the same problem taking 16’s complement of subtrahend. So let’s solve the problem by this method.

The 16’s complement of A3B₁₆

= 15’s complement of A3B₁₆ + 1

= 5C4₁₆ + 1

= 5C5₁₆

Now add ABC₁₆ and 5C5₁₆.

Digit (Starting from Right)	Remainder	Quotient
C+5=16	17%16=1	16/16=1
(Quotient added) B+C+1=24	24%16=8	24/16=1
(Quotient added) A+5+1=16	16%16=0	16/16=1
(Quotient added) 0+0+1=1	1%16=1	1/16=0

Thus the remainder is 1081₁₆. Now we have to discard the left most end carry 1 then number obtained is final value.

Hence, ABC₁₆ – A3B₁₆=81₁₆.

Question 5: Subtract the hexadecimal numbers 67A₁₆ and 549₁₆?

Solution: The solution of this problem is as follows:

By using 15’s complement of subtrahend:

15’s complement of 549₁₆ is found as:

15-5=10 (in hexadecimal number system 10=A)

15-4=11 (in hexadecimal number system 11=B)

15-9=6

So the 15’s complement is AB6₁₆.

Now add 67A₁₆ and AB6₁₆.

Digit (Starting from Right)	Remainder	Quotient
A+6=16	16%16=0	16/16=1
(Quotient added) 7+B+1=19	19%16=3	19/16=1
(Quotient added) 6+A+1=17	17%16=1	17/16=1
(Quotient added) 0+0+1=1	1%16=1	1/16=0

Thus the remainder is 1130₁₆. Now discard left most end carry 1 and add 1 to the right most digit.

Hence, 1130₁₆ become 130₁₆ and after adding 1 we get 131₁₆.

So 67A₁₆ – 549₁₆ = 131₁₆.

By using 16’s complement of subtrahend:

16’s complement of 549₁₆

= 15’s complement of 549₁₆ + 1

= AB6₁₆+1

= AB7₁₆

Now add 67A₁₆ and AB7₁₆.

Digit (Starting from Right)	Remainder	Quotient
A+7=17	17%16=1	17/16=1
(Quotient added) 7+B+1=19	19%16=3	19/16=1
(Quotient added) 6+A+1=17	17%16=1	17/16=1
(Quotient added) 0+0+1=1	1%16=1	1/16=0

Thus when we arrange the remainder we get 1131₁₆. Now discard the left most end carry and we will get the final solution.

Hence after discarding end carry of 1131₁₆.

We get 131₁₆.

Question 6: Solve ABC₁₆+FAB₁₆-2AC₁₆?

Solution: first we add ABC₁₆ and FAB₁₆ then we subtract 2AC₁₆ from the sum of ABC₁₆ and FAB₁₆.

Digit (Starting from Right)	Remainder	Quotient
C+B=23	23%16=7	23/16=1
(Quotient added) B+A+1=22	22%16=6	22/16=1
(Quotient added) A+F+1=26	26%16=10(A)	26/16=1
(Quotient added) 0+0+1=1	1%16=1	1/16=0

Thus sum of ABC₁₆ and FAB₁₆ is 1A67₁₆.

Now subtract 1A67₁₆ and 2AC₁₆.

15’s complement of 2AC is obtained as:

15-2=13 (in hexadecimal number system 13=D)

15-A=5

15-C=3

Thus the 15’s complement of 2AC₁₆ is D53₁₆.

Now add 1A67₁₆ and D53₁₆.

Digit (Starting from Right)	Remainder	Quotient
7+3=10	10%16=10(A)	10/16=0
(Quotient added) 6+5+0=11	11%16=11(B)	11/16=0
(Quotient added) A+D+0=23	23%16=7	23/16=1
(Quotient added) 1+0+1=2	2%16=2	2/16=0

Thus we get 27BA₁₆. After discarding 1 from left most digit of 27BA₁₆ and then we add 1.

Hence 27BA₁₆ becomes 17BA₁₆ and 17BA₁₆+1=17BB₁₆.

ABC₁₆+FAB₁₆-2AC₁₆ = 17BB₁₆.

Alternative Method of Addition and Subtraction of Hexadecimal Numbers

There is an alternative method also to find addition and subtraction of hexadecimal numbers. The steps are as follows:

1. Convert each hexadecimal numbers into decimal numbers.
2. Add or subtract the decimal numbers obtained from step one.
3. Convert the decimal number obtained from step two into hexadecimal number.
4. The hexadecimal number obtained from step three is the final answer.

Question 7: Add the hexadecimal numbers A21₁₆ and 2B1₁₆?

Solution: To add the above two hexadecimal numbers we have to follow the above steps:

Convert A21₁₆ and 2B1₁₆ into decimal numbers:

A21₁₆

= Ax16²+2×16¹+1×16⁰

= 10×256+2×16+1×1

= 2560+32+1

= 2593₁₀

2B1₁₆

= 2×16²+Bx16¹+1×16⁰

= 2×256+11×16+1×1

= 512+176+1

= 689₁₀

Add 2593₁₀ and 689₁₀:

2593₁₀ + 689₁₀ = 3282₁₀

Convert 3282₁₀ into hexadecimal number:

          16|3282     Remainder

          16|205             2

          16|12               13 (D)

          16|0                 12 (C)

So the sum is CD2₁₆.

Question 8: Subtract the hexadecimal numbers ABC₁₆ and A8C₁₆?

Solution: Convert the hexadecimal numbers ABC₁₆ and A8C₁₆ Into decimal numbers:

ABC₁₆

= Ax16²+Bx16¹+Cx16⁰

= 10×256+11×16+13×1

= 2560+176+12

= 2748₁₀ 

A8C₁₆

=Ax16²+8×16¹+Cx16⁰

= 10×256+8×16+12×1

= 2560+128+12

= 2700₁₀

Subtract 2749₁₀ and 2668₁₀:

2748₁₀ – 2700₁₀ = 48₁₀

Convert 48₁₀ into hexadecimal number:

          16|81          Remainder

          16|3                 0

          16|0                 3

So the answer is 30₁₆.

Multiplication and Division of Hexadecimal Numbers

The alternative method of finding multiplication or division of hexadecimal numbers is shown below:

1. Change the hexadecimal numbers into decimal numbers.
2. Multiply or divide the hexadecimal numbers.
3. The decimal number obtained in the second step has to be changed into hexadecimal number and that is the final value.

Question 9: Find the multiplication of the hexadecimal numbers 11A₁₆ and 5A3₁₆?

Solution: The solution is found by the above method as:

Change hexadecimal numbers into decimal numbers:

11A₁₆

= 1×16²+1×16¹+Ax16⁰

= 1×256+1×16+10×1

= 256+16+10

= 282₁₀

5A3₁₆

= 5×16²+Ax16¹+3×16⁰

= 5×256+10×16+3×1

= 1280+160+3

= 1443₁₀

Now multiply 282₁₀ and 1443₁₀:

282₁₀ x 1443₁₀ = 406926₁₀

Change 406926₁₀ into hexadecimal number:

          16|406926          Remainder

          16|25432                14(E)

          16|1589                   8

          16|99                       5

          16|6                         3

              |0                         6

So the multiplication of 11A₁₆ and 5A3₁₆ is 6358E_16­.

Question 10: Find division of ABC₁₆ by A1₁₆?

Solution: The solution is as follows:

Change the hexadecimal numbers into decimal numbers:

ABC₁₆

= Ax16²+Bx16¹+Cx16⁰

= 10×256+11×16+13×1

= 2560+176+12

= 2748₁₀ 

A1₁₆

= Ax16¹+1×16⁰

= 10×16+1×1

= 160+1

= 161₁₀

Divide 2748₁₀ by 161₁₀:

2748₁₀ / 161₁₀ = 17₁₀ (Around)

Now change 17₁₀ into hexadecimal number:

          16|17         Remainder

          16|1                 1

              |0                 1

So the answer is 11₁₆.

See Also:

1. Primary Secondary and Catch Memory
2. Addition, Subtraction, Multiplication and Division of Binary Numbers
3. Addition, Subtraction, Multiplication and Division of Octal Numbers