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Binary Division Example
( Enter your problem )
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- Addition
- Subtraction
- Multiplication
- Division
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Other related methods
- Addition, Subtraction, Multiplication, Division of two Decimal numbers
- Addition, Subtraction, Multiplication, Division of two Binary numbers
- Addition, Subtraction, Multiplication, Division of two Octal numbers
- Addition, Subtraction, Multiplication, Division of two Hexadecimal numbers
- Addition, Subtraction, Multiplication, Division of two any base numbers
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4. Binary Division
1. Find division of `(1011)_2` and `(10)_2`
Solution:
Solution is
| | 1 | 0 | 1 | 10 | 1 | 0 | 1 | 1 | − | 1 | 0 | | | =10 × 1 | | | | 1 | | | | | − | 0 | | =10 × 0 | | | | 1 | 1 | | | | − | 1 | 0 | =10 × 1 | | | | | 1 | |
`:.` 1011 `-:` 10= 101 Remainder 1
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Step by step solutionStep by step solution :Step-1 :Set up the problem with long division bracket. Put dividend inside bracket and divisor on outside left. Step-2 :10 goes into 10 (1-times). Put a 1 in the next place of quotient and multiply 10 by 1 to get 10. Subtract 10 from 10 to get remainder `(10-10=0)`. Step-3 :Now, bring down 1 from the dividend, to make 1 Step-4 :10 goes into 1 (0-times). Put a 0 in the next place of quotient and multiply 10 by 0 to get 0. Subtract 0 from 1 to get remainder `(1-0=1)`. | | 1 | 0 | | 10 | 1 | 0 | 1 | 1 | − | 1 | 0 | | | =10 × 1 | | | | 1 | | | | | − | 0 | | =10 × 0 | | | | 1 | | | Step-5 :Now, bring down 1 from the dividend, to make 11 | | 1 | 0 | | 10 | 1 | 0 | 1 | 1 | − | 1 | 0 | | | =10 × 1 | | | | 1 | | | | | − | 0 | | =10 × 0 | | | | 1 | 1 | | Step-6 :10 goes into 11 (1-times). Put a 1 in the next place of quotient and multiply 10 by 1 to get 10. Subtract 10 from 11 to get remainder `(11-10=1)`. | | 1 | 0 | 1 | 10 | 1 | 0 | 1 | 1 | − | 1 | 0 | | | =10 × 1 | | | | 1 | | | | | − | 0 | | =10 × 0 | | | | 1 | 1 | | | | − | 1 | 0 | =10 × 1 | | | | | 1 | |
2. Find division of `(11101)_2` and `(101)_2`
Solution:
Solution is
| | | 1 | 0 | 1 | 101 | 1 | 1 | 1 | 0 | 1 | − | 1 | 0 | 1 | | | =101 × 1 | | | 1 | 0 | 0 | | | | | | − | 0 | | =101 × 0 | | | 1 | 0 | 0 | 1 | | | | − | 1 | 0 | 1 | =101 × 1 | | | | 1 | 0 | 0 | |
`:.` 11101 `-:` 101= 101 Remainder 100
| | 101 table | 101 | × | 1 | = | 101 | 101 | × | 10 | = | 1010 |
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Step by step solutionStep by step solution :Step-1 :Set up the problem with long division bracket. Put dividend inside bracket and divisor on outside left. Step-2 :101 goes into 111 (1-times). Put a 1 in the next place of quotient and multiply 101 by 1 to get 101. Subtract 101 from 111 to get remainder `(111-101=10)`. | | | 1 | | | 101 | 1 | 1 | 1 | 0 | 1 | − | 1 | 0 | 1 | | | =101 × 1 | | | 1 | 0 | | | | Step-3 :Now, bring down 0 from the dividend, to make 100 | | | 1 | | | 101 | 1 | 1 | 1 | 0 | 1 | − | 1 | 0 | 1 | | | =101 × 1 | | | 1 | 0 | 0 | | | Step-4 :101 goes into 100 (0-times). Put a 0 in the next place of quotient and multiply 101 by 0 to get 0. Subtract 0 from 100 to get remainder `(100-0=100)`. | | | 1 | 0 | | 101 | 1 | 1 | 1 | 0 | 1 | − | 1 | 0 | 1 | | | =101 × 1 | | | 1 | 0 | 0 | | | | | | − | 0 | | =101 × 0 | | | 1 | 0 | 0 | | | Step-5 :Now, bring down 1 from the dividend, to make 1001 | | | 1 | 0 | | 101 | 1 | 1 | 1 | 0 | 1 | − | 1 | 0 | 1 | | | =101 × 1 | | | 1 | 0 | 0 | | | | | | − | 0 | | =101 × 0 | | | 1 | 0 | 0 | 1 | | Step-6 :101 goes into 1001 (1-times). Put a 1 in the next place of quotient and multiply 101 by 1 to get 101. Subtract 101 from 1001 to get remainder `(1001-101=100)`. | | | 1 | 0 | 1 | 101 | 1 | 1 | 1 | 0 | 1 | − | 1 | 0 | 1 | | | =101 × 1 | | | 1 | 0 | 0 | | | | | | − | 0 | | =101 × 0 | | | 1 | 0 | 0 | 1 | | | | − | 1 | 0 | 1 | =101 × 1 | | | | 1 | 0 | 0 | |
3. Find division of `(11101)_2` and `(110)_2`
Solution:
Solution is
| | | 1 | 0 | 0 | 110 | 1 | 1 | 1 | 0 | 1 | − | 1 | 1 | 0 | | | =110 × 1 | | | | 1 | 0 | | | | | | − | 0 | | =110 × 0 | | | | 1 | 0 | 1 | | | | | | − | 0 | =110 × 0 | | | | 1 | 0 | 1 | |
`:.` 11101 `-:` 110= 100 Remainder 101
| | 110 table | 110 | × | 1 | = | 110 | 110 | × | 10 | = | 1100 |
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Step by step solutionStep by step solution :Step-1 :Set up the problem with long division bracket. Put dividend inside bracket and divisor on outside left. Step-2 :110 goes into 111 (1-times). Put a 1 in the next place of quotient and multiply 110 by 1 to get 110. Subtract 110 from 111 to get remainder `(111-110=1)`. | | | 1 | | | 110 | 1 | 1 | 1 | 0 | 1 | − | 1 | 1 | 0 | | | =110 × 1 | | | | 1 | | | | Step-3 :Now, bring down 0 from the dividend, to make 10 | | | 1 | | | 110 | 1 | 1 | 1 | 0 | 1 | − | 1 | 1 | 0 | | | =110 × 1 | | | | 1 | 0 | | | Step-4 :110 goes into 10 (0-times). Put a 0 in the next place of quotient and multiply 110 by 0 to get 0. Subtract 0 from 10 to get remainder `(10-0=10)`. | | | 1 | 0 | | 110 | 1 | 1 | 1 | 0 | 1 | − | 1 | 1 | 0 | | | =110 × 1 | | | | 1 | 0 | | | | | | − | 0 | | =110 × 0 | | | | 1 | 0 | | | Step-5 :Now, bring down 1 from the dividend, to make 101 | | | 1 | 0 | | 110 | 1 | 1 | 1 | 0 | 1 | − | 1 | 1 | 0 | | | =110 × 1 | | | | 1 | 0 | | | | | | − | 0 | | =110 × 0 | | | | 1 | 0 | 1 | | Step-6 :110 goes into 101 (0-times). Put a 0 in the next place of quotient and multiply 110 by 0 to get 0. Subtract 0 from 101 to get remainder `(101-0=101)`. | | | 1 | 0 | 0 | 110 | 1 | 1 | 1 | 0 | 1 | − | 1 | 1 | 0 | | | =110 × 1 | | | | 1 | 0 | | | | | | − | 0 | | =110 × 0 | | | | 1 | 0 | 1 | | | | | | − | 0 | =110 × 0 | | | | 1 | 0 | 1 | |
This material is intended as a summary. Use your textbook for detail explanation. Any bug, improvement, feedback then
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