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You need 20 bits of address bus to address 1 Mb of memory.
When the low order bits of the address are used to select the memory bank it is interleaved.
It requires 6 bits to address 64 words. It does not matter what the word size is.
The data bus in the 8086 is 16 bits in size, while the address bus is 20 (16bits would only address 64KB of memory, an extra 4 bits allows to address the total of 1MB, this is done trough segmentation of the memory). To form a multiplexed of data bus and address bus, four bits of 8086 address bus are grounded.
8 bits
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.
15 bits
This is due to the fact that 16bits would only address 64KB of memory, which even then was very little. The answer for this was to come up with an extra 4 bits to address the total 1MB, this is done trough segmentation of the memory. Google it.
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
8086 has 20 address lines. Therefore it can address 220 bits or 1,048,576 bits of memory, or roughly 1 MB (mega byte).