Question: odd parity os Original number = 1010
0

Original number = 1010

Odd parity

For 4-bit, 3 parity bits $p_1, p_2, p_3$ are appended in locations 1,2,4 respectively.

Position 1 2 3 4 5 6 7
- P1 P2 D1 P3 D2 D3 D4
original number - - 1 - 0 1 0
odd parity in positions 1.3.5.7 requires P1 = 0 0 - 1 - 0 1 0
odd parity in positions 2,3,6,7 requires P2 = 1 0 1 1 - 0 1 0
odd parity in positions 4,5,6,7 requires P3 = 0 0 1 1 0 0 1 0

$\therefore$ Hamming code for 1010 with odd parity os 0110010

$(-89)_{10}$

Binary :

$\therefore$ $(89)_{10}$

$= (01011001)_2$

1. signed magnitude

since, the number is negative, sign bit = 1

magnitude = 01011001

$\therefore$ $(-89)_{10} = (101011001)_2$

2. 1's complement,

$(-89)_{10} = (10100110)_2$

1's complement of $(-89)_{10}$ = 10100110

3. 2's complement,

2's complement = 1's complement + 1

$\therefore$ 2's complement of $(-89)_{10} = 10100111$

$(BC5)_H - (A2B)_H$

15's complement of A2B,

$\therefore$ 16's complement of A2B = 5D4 + 1 = 5D5