Answer:
a. 10bits
b. 7 bits
c. 6 bits
Step-by-step explanation:
a. for 0 to 512
# of numbers = 512 - 0 + 1 =513
[log ₂513] = 9 bits
we actually need 10 bits
b. for 0 to 75
# of numbers = 75 - 0 + 1 =76
[log ₂76] = 6 bits
we actually need 7 bits
c. for -20 to 13
# of numbers = 13 - (-20) + 1 =34
[log ₂34] = 5 bits
we actually need 6 bits