For 983 and 187, Find Excess 3 Subtraction using 9's complement
Solution:
`983-187` Excess 3 subtraction using `9's` complement
Steps for Excess 3 subtraction using `9's` complement
For `A-B`
1. Take `9's` complement for B
2. Add it to A using binary addition
3. If carry then add it to the next bits and consider this bits as Invalid
4. Add 3 for Invalid bits and subtract 3 for others
5. In final result, if carry is occured then add it the remaining result and if
there is no any carry over, then take `9's` complement of the result and it is negative.
1. Take `9's` complement for `187`
Note : 9's complement of a number is obtained by subtracting all bits from 999
9's complement of 187 is
2. Add `983` and `812` using Excess 3 addition
| Excess 3 code for 983 : | 1100 | 1011 | 0110 |
| Excess 3 code for 812 : | 1011 | 0100 | 0101 |
|
| Addition : | 10111 | 1111 | 1011 |
| If carry generated then add 3 otherwise subtract 3 : | -0011 | -0011 | -0011 |
|
| Result : | 10100 | 1100 | 1000 |
| Decimal value : | 20 | 12 | 8 |
| Excess 3 value : | 17 | 9 | 5 |
The left most bit of the result is 1, called carry and This will be added to `795`.
`795+1=796`
So final answer of Excess 3 Subtraction is `796`
This material is intended as a summary. Use your textbook for detail explanation.
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