suppose there is a computer with 18 bits address space and the cell size is 8 bits, then what is the smallest and highest address?

Question

suppose there is a computer with 18 bits address space and the cell size is 8 bits, then

1. What is the smallest and highest address?
2. What will be the possible largest memory size of this computer in bytes, kilobytes and megabytes?

Answer 1

What is the smallest and highest address?

smallest Address = 0x0000 (HEX)
highest address = 2^18 / 8
                = 32768 = 0x8000 (HEX)

What will be the possible largest memory size of this computer in bytes, kilobytes and megabytes?

Memory Size  = 2^18 X 8 bits
Memory Size  = 262,144B   (Bytes)
Memory Size  = 262,144/1024 KB = 256 KB

Memory Size  = 256/1024 MB = 0.25 MB

Note::

For more convenient representation the units KiB and MiB are used: in that Case The memory size is 2048KiB OR 2MiB