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If the page size is 8kb, then 14 bits of the virtual address are used for the final offset. That leaves (48-14) or 34 bits to select the page table entry. That is 17,179,869,184 entries in a maximally filled page table. The physical address size has no bearing to this particular question, because you asked how many entries, not how large the page table would be.

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What are the implications of having a machine with 48-bit virtual addresses and 32-bit physical addresses?

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