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R’s AND (R1)’s COMPLEMENT OF NUMBER SYSTEMS
R’s AND (R1)’s COMPLEMENT OF BINARY NUMBER SYSTEM
= 0100111_{2}
R’s AND (R1)’s COMPLEMENT OF BASE 3 NUMBER SYSTEM
R’s AND (R1)’s COMPLEMENT OF BASE 4 NUMBER SYSTEM
R’s AND (R1)’s COMPLEMENT OF BASE 5 NUMBER SYSTEM
R’s AND (R1)’s COMPLEMENT OF BASE 6 NUMBER SYSTEM
R’s AND (R1)’s COMPLEMENT OF BASE 7 NUMBER SYSTEM
R’s AND (R1)’s COMPLEMENT OF OCTAL NUMBER SYSTEM
R’s AND (R1)’s COMPLEMENT OF BASE 9 NUMBER SYSTEM
R’s AND (R1)’s COMPLEMENT OF DECIMAL NUMBER SYSTEM
R’s AND (R1)’s COMPLEMENT OF BASE 11 NUMBER SYSTEM
R’s AND (R1)’s COMPLEMENT OF BASE 12 NUMBER SYSTEM
R’s AND (R1)’s COMPLEMENT OF BASE 13 NUMBER SYSTEM
R’s AND (R1)’s COMPLEMENT OF BASE 14 NUMBER SYSTEM
R’s AND (R1)’s COMPLEMENT OF BASE 15 NUMBER SYSTEM
R’s AND (R1)’s COMPLEMENT OF HEXADECIMAL NUMBER SYSTEM
HEXADECIMAL ARITHMETIC
ADDITION OF HEXADECIMAL
The addition of hexadecimal numbers are achieved by following the given two steps given below.
 Add the right most digits of each hexadecimal numbers.
 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.
Example 1: Find the sum of the numbers A23_{16} and 1B1_{16}?
Solution: The addition of A23_{16} and 1B1_{16} 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)  11/16=0 
Thus the addition of A23_{16} + 1B1_{16} = BD4_{16}.
Example 2: Find the sum of the hexadecimal numbers 2AC_{16} and FAB_{16}?
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 2AC_{16} and FAB_{16} is 1257_{16}.
Example 3: Find the addition of A2B.D4_{16} and 247.65_{16}?
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 C73.39_{16}.
SUBTRACTION OF HEXADECIMAL
Subtraction of hexadecimal numbers is obtained by following the given steps:
 First find 15’s complement or 16’s of the subtrahend.
 Add minuend and 15’s complement or 16’s of the subtrahend using the above addition steps of hexadecimal numbers.
 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.
Example 1: Subtract the hexadecimal numbers ABC_{16} and A3B_{16}?
Solution: here we will find the subtraction of the two given numbers using 15’s complement. So the solution for 15’s complement of A3B_{16} is as follows.
15 – A = 5
15 – 3 = 12 (in hexadecimal number system 12=C)
15 – B = 4
So the 15’s complement of A3B_{16} is 5C4_{16}.
Now add ABC_{16}and 5C4_{16}.
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 1080_{16}. Now we have to discard the left most end carry 1 and then we have to add 1 to the right most digit.
Thus 1080_{16}becomes 80_{16}+1=81_{16
}Hence, ABC_{16}– A3B_{16}=81_{16}.
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 A3B_{16
}= 15’s complement of A3B_{16} + 1
= 5C4_{16} + 1
= 5C5_{16
}Now add ABC_{16}and 5C5_{16}.
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 1081_{16}. Now we have to discard the left most end carry 1 then number obtained is final value.
Hence, ABC_{16} – A3B_{16} = 81_{16}.
Example 2: Subtract the hexadecimal numbers 67A_{16} and 549_{16}?
Solution: The solution of this problem is as follows:
By using 15’s complement of subtrahend:
15’s complement of 549_{16} 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 AB6_{16}.
Now add 67A_{16}and AB6_{16}.
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 1130_{16}. Now discard left most end carry 1 and add 1 to the right most digit.
Hence, 1130_{16}become 130_{16} and after adding 1 we get 131_{16}.
So 67A_{16} – 549_{16} = 131_{16}.
By using 16’s complement of subtrahend:
16’s complement of 549_{16}
= 15’s complement of 549_{16} + 1
= AB6_{16} + 1
= AB7_{16
}Now add 67A_{16} and AB7_{16}.
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 1131_{16}. Now discard the left most end carry and we will get the final solution.
Hence after discarding end carry of 1131_{16}.
We get 131_{16}.
Example 3: Solve ABC_{16} + FAB_{16} – 2AC_{16}?
Solution: first we add ABC_{16} and FAB_{16} then we subtract 2AC_{16}from the sum of ABC_{16} and FAB_{16}.
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 ABC_{16}and FAB_{16} is 1A67_{16}.
Now subtract 1A67_{16} and 2AC_{16}.
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 2AC_{16} is D53_{16}.
Now add 1A67_{16}and D53_{16}.
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 27BA_{16}. After discarding 1 from left most digit of 27BA_{16}and then we add 1.
Hence 27BA_{16}becomes 17BA_{16} and 17BA_{16}+1=17BB_{16}.
ABC_{16} + FAB_{16} – 2AC_{16} = 17BB_{16}.
ALTERNATIVE METHOD FOR 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:
 Convert each hexadecimal numbers into decimal numbers.
 Add or subtract the decimal numbers obtained from step one.
 Convert the decimal number obtained from step two into hexadecimal number.
 The hexadecimal number obtained from step three is the final answer.
Example 1: Add the hexadecimal numbers A21_{16} and 2B1_{16}?
Solution: To add the above two hexadecimal numbers we have to follow the above steps:
Convert A21_{16} and 2B1_{16 }into decimal numbers:
A21_{16
}= Ax16^{2} + 2×16^{1} + 1×16^{0
}= 10×256 + 2×16 + 1×1
= 2560 + 32 + 1
= 2593_{10
}
2B1_{16
}= 2×16^{2} + Bx16^{1} + 1×16^{0
}= 2×256 + 11×16 + 1×1
= 512 + 176 + 1
= 689_{10}
Add 2593_{10} and 689_{10}.
2593_{10} + 689_{10} = 3282_{10
}Convert 3282_{10} into hexadecimal number:
163282 Remainder
16205 2
1612 13 (in hexadecimal number system 13=D)
160 12 (in hexadecimal number system 12=C)
So the sum is CD2_{16}.
Example 2: Subtract the numbers ABC_{16} and A8C_{16}?
Solution: Convert the hexadecimal numbers ABC_{16} and A8C_{16} Into decimal numbers:
ABC_{16
}= Ax16^{2} + Bx16^{1} + Cx16^{0
}= 10×256 + 11×16 + 13×1
= 2560 + 176 + 12
= 2748_{10}
A8C_{16}
= Ax16^{2} + 8×16^{1} + Cx16^{0}
= 10×256 + 8×16 + 12×1
= 2560 + 128 + 12
= 2700_{10}
Subtract 2749_{10 }and 2668_{10}:
2748_{10} – 2700_{10} = 48_{10
}Convert 48_{10} into hexadecimal number:
1681 Remainder
163 0
0 3
So the answer is 30_{16}.
MULTIPLICATION AND DIVISION OF HEXADECIMAL NUMBERS
The alternative method of finding multiplication or division of hexadecimal numbers is shown below:
 Change the hexadecimal numbers into decimal numbers.
 Multiply or divide the hexadecimal numbers.
 The decimal number obtained in the second step has to be changed into hexadecimal number and that is the final value.
Example 1: Find the multiplication of the hexadecimal numbers 11A_{16} and 5A3_{16}?
Solution: The solution is found by the above method as:
Change hexadecimal numbers into decimal numbers:
11A_{16
}= 1×16^{2} + 1×16^{1} + Ax16^{0
}= 1×256 + 1×16 + 10×1
= 256 + 16 + 10
= 282_{10}
5A3_{16
}= 5×16^{2} + Ax16^{1} + 3×16^{0
}= 5×256 + 10×16 + 3×1
= 1280 + 160 + 3
= 1443_{10}
Now multiply 282_{10} and 1443_{10}:
282_{10} x 1443_{10} = 406926_{10
}Change 406926_{10} into hexadecimal number:
16406926 Remainder
1625432 14(in hexadecimal number system 14=E)
161589 8
1699 5
166 3
0 6
So the multiplication of 11A_{16} and 5A3_{16} is 6358E_{16}.
Example 2: Find division of ABC_{16} by A1_{16}?
Solution: The solution is as follows:
Change the hexadecimal numbers into decimal numbers:
ABC_{16
}= Ax16^{2} + Bx16^{1} + Cx16^{0
}= 10×256 + 11×16 + 13×1
= 2560 + 176 + 12
= 2748_{10}
A1_{16}
= Ax16^{1} + 1×16^{0
}= 10×16 + 1×1
= 160 + 1
= 161_{10
}Divide 2748_{10} by 161_{10}:
2748_{10} / 161_{10} = 17_{10} (Around)
Now change 17_{10 }into hexadecimal number:
1617 Remainder
161 1
0 1
So the answer is 11_{16}.
15’S AND 16’S COMPLEMENTS OF HEXADECIMAL NUMBERS
15’S AND 16’S COMPLEMENTS OF HEXADECIMAL NUMBERS
general we can say if a number system has its radix or base R then only R’s and (R1)’s complements can be found of that number. The general formula for finding R’s complement and (R1)’s complement are shown below:
 R’s complement of any number system = (R^{n})_{10}N and
 (R1)’s complement of any number system = {(R^{n})_{10}1)}N
Where R = radix or base of that number system
15’S COMPLEMENT OF HEXADECIMAL NUMBERS
 Calculate (R^{n})_{10}1.
 Convert the hexadecimal number N into decimal number.
 Subtract the values obtained from above two steps with each other.
 Convert the decimal value obtained in step3 into hexadecimal.
SHORTCUT METHOD FOR FINDING 15’S COMPLEMENT OF HEXADECIMAL NUMBERS
16’S COMPLEMENT OF HEXADECIMAL NUMBERS
SHORTCUT METHOD FOR FINDING 16’S COMPLEMENT OF HEXADECIMAL NUMBERS
SEE ALSO:
7’S AND 8’S COMPLEMENT OF OCTAL NUMBERS
1’S AND 2’S COMPLEMENT OF BINARY NUMBER AND SIGNED NUMBERS
7’s AND 8’s COMPLEMENT OF OCTAL NUMBER
7’s COMPLEMENT OF OCTAL NUMBER
To find 7’s complement or (r – 1)’s complement of Octal number we have to follow some steps. There is a general formula to find (r1)’s complement of any number system. So
(r1)’s complement = {(r^{n})_{10} – 1} – N
r = Radix or Base
n = Number of digits
N = Number.
 Calculate (r^{n})_{10} – 1.
 Convert the given Octal number (N) in Decimal system.
 Subtract the values obtained from Step One and Step Two.
 The decimal value obtained from Step Three has to be converted into octal value and that will be 7’s complement of octal number.Let us have an example to have a better consideration of 7’s complement of octal number.
OCTAL ARITHMETIC
The addition and subtraction of octal number are shown here. An alternative method of solving this types of problem is also discussed in this section below.
OCTAL ADDITION
The addition of two or more octal numbers can be achieved by following some specific rules which are listed below. Since there are eight digits in this system starting from 0 and the last digit are 7. The additions of octal numbers are discussed below.
 Add the right most digits of each octal numbers.
 Find the modulo of the sum of digits means divide the sum by 8 and the remainder so obtained is the octal equivalent of the sum and the quotient is added to the next sum of digits from right side. Perform the STEP 2 until we add the left most digits.
Let us take an example to have a better idea how the whole operation takes place.
Exp1: Add the octal numbers (123)_{8} and (527)_{8}.
Digit (Starting from Right)  Remainder  Quotient 
3+7=10  10%8=2  10/8=1 
(Quotient added) 1+2+2=5  5%8=5  5/8=0 
(Quotient added) 0+1+5=6  6%8=6  6/8=0 
Hence, the sum of (123)_{8} and (527)_{8 }is (652)_{8}.
[NOTE: WHEN THE SUM OF THE DIGITS IS LESS THAN 8 THEN THE MODULO OF THE SUM WILL BE THE SUM ITSELF AND THE QUOTIENT WILL BE ZERO.]
Exp2: Add the octal numbers (325.4)_{8} and (723.8)_{8}.
Digits (Starting from Right)  Remainder (Octal Equivalent)  Quotient 
4+8 = 12  12%8 = 4  12/8 = 1 
(Quotient Added) 1+5+3 = 9  9%8 = 1  9/8 = 1 
(Quotient Added) 1+2+2 = 5  5%8 = 5  5/8 = 0 
(Quotient Added) 0+3+7 = 10  10%8 = 2  10/8 = 1 
(Quotient Added) 1 = 1  1%8 = 1  1/8 = 0 
Therefore, the sum is (1251.4)_{8}.
OCTAL SUBTRACTION
Octal subtraction is performed as we perform the binary subtraction. In binary subtraction, first we find 1’s complement or 2’s complement of subtrahend in the same way first we will find 7’s complement or 8’s complement of that octal number then we add the minuend and the subtrahend (The addition is performed as we have stated above.). So to find the subtraction of two octal numbers we have to follow two steps:
 Find either 7’s complement or 8’s of the subtrahend.
 Add the minuend and the subtrahend.
By using 7’s complement
Let us take an example:
Exp1: subtract (453)_{8} – (234)_{8}.
STEP 1: 7’s complement of (234)_{8}= 777 – 234 = (543)_{8
}STEP 2: Add the numbers (453)_{8 }and (543)_{8}.
Digits (Starting from Right)  Remainder (Octal Equivalent)  Quotient 
3+3 = 6  6%8 = 6  6/8 = 0 
(Quotient added) 0+5+4 = 9  9%8 = 1  9/8 = 1 
(Quotient added) 1+4+5 = 10  10%8 = 2  10/8 = 1 
(Quotient added) 1 = 1  1%8 = 1  1/8 = 0 
So, the sum is (1216)_{8.} Now, discard the end carry and add 1 to the right most digit and that will be the answer.
Hence, (453)_{8} – (234)_{8} = (217)_{8}.
Exp2: Subtract (165)_{8 }– (76)_{8}.
STEP 1: 7’s complement of (76)_{8 }= 77 – 76 = (1)_{8 }STEP 2: Add the octal numbers (165)_{8 }and (1)_{8}.
Digits (Starting from Right)  Remainder (Octal Equivalent)  Quotient 
1+5 = 6  6%8 = 6  6/8 = 0 
(Quotient added)0+6 = 6  6%8 = 6  9/8 = 0 
(Quotient added)0+1 = 1  1%8 = 1  1/8 = 0 
The sum is (166)_{8}.
Now, discard the end carry and add 1 to the right most digit.
Hence, (165)_{8} – (76)_{8} = (67)_{8}.
By 8’s complement:
Exp1: Find the value of (574)_{8}– (341)_{8}.
To solve this problem we use 8’s complement of subtrahend. There are also two steps first find the 8’s complement of subtrahend and the follow the STEP 2 of the 7’s complement method.
Now, 8’s complement any octal number = 7’s complement of that octal number + 1
STEP 1: Therefore, 8’s complement of (341)_{8} = (777 – 341) + 1 = (437)_{8
}STEP 2: Add the minuend (574)_{8}and 8’s complement of subtrahend (437)_{8}.
Digits (starting from right)  Remainder (Octal Equivalent)  Quotient 
4+7 = 11  11%8 = 3  6/8 = 1 
(Quotient Added) 1+7+3 = 11  11%8 = 3  11/8 = 1 
(Quotient Added) 1+5+4 = 10  10%8 = 2  10/8 = 1

(Quotient Added) 1 = 1  1%8 = 1  1/8 = 0 
Sum of (574)_{8} and (437)_{8 }is (1233)_{8}. So, the required value is obtained by discarding the end carry only.
Therefore, (574)_{8} – (341)_{8} = (233)_{8
}Exp2: (764)_{8} – (444)_{8 }solve the problem using 8’s complement?
STEP 1: 8’s complement of subtrahend = (777 – 444) + 1 = (334)_{8
}STEP 2: Add the numbers (764)_{8 }and (334)_{8}.
Digits (Starting from Right)  Remainder (Octal Equivalent)  Quotient 
4+4=8  8%8=0  8/8=1 
(Quotient added) 1+6+3=10  10%8=2  10/8=1 
(Quotient added) 1+7+3 = 11  11%8 = 3  11/8 = 1 
(Quotient added) 1 = 1  1%8 = 1  1/8 = 0 
So the sum is (1320)_{8}.
Now discard the end carry and that will be the required answer.
Hence, (764)_{8} – (444)_{8} = (320)_{8}.
PROBLEM INCLUDING BOTH ADDITION AND SUBTRACTION OF OCTAL NUMBER
Exp1: Solve the given problem: (223)_{8}+ (527)_{8} – (436)_{8}
To solve this problem, first we are going to add the first two octal numbers then we subtract the third octal number from the sum of the first two octal numbers. So, let us start to add first two octal numbers. To add these numbers we will follow the above rules.
Digits (Starting from Right)  Remainder (Octal Equivalent)  Quotient 
3+7=10  10%8=2  10/8=1 
(Quotient added) 1+2+2=5  5%8=5  5/8=0 
(Quotient added) 0+2+5 = 7  7%8 = 7  7/8 = 0 
So the sum is (752)_{8}.
Now, find the either 7’s complement or 8’s complement of the subtrahend (436)_{8}.
Therefore, 7’s complement of (436)_{8} = 777 – 436 = (341)_{8
}Now, add the sum (752)_{8 }and (341)_{8}.
Digits (Starting from Right)  Remainder (Octal Equivalent)  Quotient 
1+2 = 3  3%8 = 3  8/8 = 0 
(Quotient added) 0+5+4 = 9  9%8 = 1  9/8 = 1 
(Quotient added) 1+7+3 = 11  11%8 = 3  11/8 = 1 
(Quotient added) 1 = 1  1%8 = 1  1/8 = 0 
So, the sum of (752)_{8 }and (341)_{8} is (1313)_{8}. Now discard the end carry and add 1 to the right most digit.
Hence, (223)_{8} + (527)_{8}– (436)_{8} = (314)_{8}.
ALTERNATE METHOD FOR ADDITION AND SUBTRACTION OF OCTAL NUMBER
The addition or subtraction of two or more octal numbers can be performed by an alternate method that follows as:
 The addition or subtraction is performed first by changing the given octal numbers into their equivalent decimal number and then we add or subtract them.
 Change the obtained final addition value or subtraction value into octal number.
Now the question arises here how an octal number can be altered in decimal number the answer is as follows:
 Multiply each digit of the octal number with its corresponding position octal value and it start from left side as for first digit 8^{0}=1, 8^{1}=8, 8^{2}=64, 8^{3}=512 and so on.
 Add all the numbers obtained from first step and that is the final decimal equivalent value of an octal number.
Let us verify the above problem by this rule.
Example: Add the octal numbers (123)_{8} and (527)_{8}.
Solution: First we change both the octal number into decimal number by above rules.
(123)_{8} = 1×8^{2}+2×8^{1}+3×8^{0
} = 64+16+3
= 83
= (83)_{10}
(527)_{8} = 5×8^{2}+2×8^{1}+7×8^{0
} = 320+16+7
= 343
= (343)_{10
}(83)_{10 }+ (343)_{10} = (426)_{10}
Now the step one has been executed. Thus we will start second step as: In the first step we have found the final decimal value, now we are going to change this decimal number into its equivalent octal number. The process of altering a decimal number into octal number is done by repeated division of the decimal number by 8 until the number becomes less than 8. The remainder found in each division together forms the octal number. The remainders are taken in order from bottom to up. Thus the octal equivalent of (426)_{10} is as obtained:
426/8=53+ 2R
53/8=6+ 5R
6/8=0+ 6R
Hence the octal equivalent is (652)_{8}. And this is the final answer in octal number system.
(123)_{8} + (527)_{8} = (652)_{8}
Similarly any octal numbers can be added by this method. Now we will discuss about the subtraction of octal number in the same procedure. The subtraction of octal numbers are illustrated below with proper example.
Example: Subtract the octal number (453)_{8} and (234)_{8}.
Solution: The subtraction process will be processed in the same way as the addition is performed above. Now we change the octal numbers (453)_{8} and (234)_{8} into decimal numbers.
(453)_{8} = 4×8^{2}+5×8^{1}+3×8^{0
} = 256+40+3
= 299
= (299)_{10}
(234)_{8} = 2×8^{2}+3×8^{1}+4×8^{0
} = 128+24+4
= 156
= (156)_{10}
(299)_{10 }– (156)_{10} = (143)_{10}
Since the first step is complete now so we will go for the next step to change the decimal value into octal value.
143/8 = 17+ 7R
17/8 = 2+ 1R
2/8 = 0+ 2R
The octal equivalence of (143)_{10} is (217)_{8}.
Hence the result is (453)_{8} – (234)_{8}= (217)_{8}.
MULTIPLICATION AND DIVISION OF OCTAL NUMBER
Multiplication and Division of octal number can also be achieved by this method easily. To take a good understanding let us go through an example:
Example: Multiply the octal number (488)_{8} and (555)_{8}.
Solution: let us change the numbers into decimal number system.
(488)_{8} = 4×8^{2}+8×8^{1}+8×8^{0
} = 256+64+8
= 328
= (328)_{10}
(555)_{8} = 5×8^{2}+5×8^{1}+5×8^{0
} = 320+40+5
= 365
= (365)_{10}
(328)_{10 }X (365)_{10} = (119720)_{10}
Now change (119720)_{10} into octal equivalent.
119720/8 = 14965+0R
14965/8 = 1870+ 5R
1870/8 = 233+ 6R
233/8 = 29+ 1R
29/8 = 3+ 5R
3/8 = 0+ 3R
Hence the octal equivalent is (351650)_{8}.
Therefore (488)_{8} X (555)_{8} = (351650)_{8}
Binary Addition Subtraction Multiplication and Division
Binary Addition
POSITIONAL NUMBER SYSTEM
In a positional number system, there are only a few called digits, and these digits represent different values depending on the position they occupy in the number. The value of each digit in such a number is determined by three consideration.
 The digit itself.
 The position of the digit in the number.
 The base of the number system.