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With 64 bit virtual addresses, a 4KB page and 256 MB of RAM, an inverted page table requires:

  1. 8192 entries
  2. 16384 entries
  3. 32768 entries
  4. 65536 entries

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d) answer is 65536

size of ram / page size = inverted page table entries 

Inverted page tables have 1 entry / physical frame of memory

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