Memory called RDRAM that conforms to the RIMM form factor.
16-bit memory
SIMM
GREGS
DRAM
Real mode uses 16 bit data path while protected mode uses 32 bit data path.
The 8088 brings in data in 8 bit chunks form the memory and must bring in two because internally it is 16 bit. The 8086 brings in the full 16 bits at one time. so it is the path to memory that differs.
The Intel 8085 microprocessor uses an 8-bit data bus and a 16-bit address bus. It can address 64 KB of memory, with each memory location capable of storing a byte. Therefore, the maximum positive number that can be represented in an 8-bit register is 255 (2^8 - 1), while in a 16-bit address space, the maximum is 65,535 (2^16 - 1) when considering the entire memory range.
If you assume that it has a 16-bit data bus, then it would be 128k so the microprocessor can access 2^16 points, which is 64k (from it being a 16bit address) 16bits = 2 bytes (memory) so through a 16 bit memory, it can access 2*64k, which is 128k alternatively, if its 8bit memory, 8bits=1byte 1*64k = 64k I'm no expert, and i was searching for the answer myself, hope this helped
16 bit without ECC and 18 bit with ECC Source : A+ Guide to hardware 4e
RIMM
ip address
16-bit real mode is a CPU operating mode used primarily in x86 architecture, allowing the processor to access memory and execute instructions in a 16-bit environment. In this mode, the CPU operates with a memory address space limited to 1 MB and uses 16-bit segment and offset addressing. Real mode is the default mode upon system startup, enabling compatibility with older software and operating systems like MS-DOS. However, it lacks advanced features such as memory protection and multitasking found in protected mode.