Why is the 4mb = 2^22 instead of 4*2^20 what am I missing here and the same for 32mb = 2^25 how do we determine the power ?

2- Suppose a processor of the Harvard architecture has 4MB instruction memory and 32MB data memory. If the memory address bus is shared between instruction memory and data memory, design a memory allocation scheme for this processor (i.e. address range for the instruction memory and the data memory). Solution: 4MB 22 bytes, Address range Ox0000_0000 -0x003F_FFFF 32MB = 225 bytes, Address range 0x0000 0000-0x01FFFFFF There are many ways to divide the space. For example: Instruction memory: 0x0010 0000-ox013F FFFF Data Memory: 0x2000_0000 0x21FF_FFFF