You need 20 bits of address bus to address 1 Mb of memory.
15 bits
You need 12 address lines to access 4K of memory. 212 = 4096. log2 (4096) = 12.
That depends on the memory architecture of the system.if the memory chips are byte wide and not used to create a multibyte bus, 11 address bits are needed.if the memory chips are 32 bits wide, 9 address bits are needed (with the CPU internally selecting which of the 4 bytes it will use).it the memory chips are 64 bits wide, 8 address bits are needed (with the CPU internally selecting which of the 8 bytes it will use.if the memory chips are 4 bits wide, 12 address bits will be needed and the CPU must perform 2 memory cycles per byte that it needs. (yes, I have seen a computer that worked this way!)etc.
When the low order bits of the address are used to select the memory bank it is interleaved.
4K (4096) of addressable space is defined by 12 bits of address space, because 212 = 4096.
4 pages -> 2^2 bits 1024 bytes -> 2^10 bits 64 frames -> 2^6 bits Therefore: Logical memory = 2+10=12 bits Physical memory = 10 +6 =16 bits
8 bits
It takes 23 address lines to address 8 mb of memory.
16384
It requires 30 address bits to address 1GB of RAM.230 = 1,073,741,824
With a 20-bit address bus, a computer can address approximately 1,048,576 memory locations, which is equivalent to 1 megabyte of memory.