9
21 bits.
24 bits are needed for the program counter. Assuming the instructions are 32 bits, then 32 bits are needed for the instruction register.
There are 8 bits
8 bits if unsigned, 9 bits if signed
11 bits. 211 = 2048
how many bits are needed to represent decimal values ranging from 0 to 12,500?
Every bit can be either 0 or 1. Therefore 4 bits can encode a maximum of 42 = 16 digits.
5
1
18 in binary is 10010 Since 18 can't be written in term of 2 to the power x, the number of bits needed is 5. The answer is 5
1200
Count them: 643(10)=1010000011(2)