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Editorial Team
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Editorial Team
Asked: May 18, 20262026-05-18T09:15:40+00:00 2026-05-18T09:15:40+00:00

Sorry this is a really silly question, was just looking through an old exam

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Sorry this is a really silly question, was just looking through an old exam paper for my Operating Systems class and noticed this question can come up every so often. How does one work out the answer to this ? I don’t really understand.

Another question is :

What are the maximum and minimum unsigned integers that can be represented by 7 bits, 9 bits and 10 bits?

Would anyone be able to tell me how you would work this out ? Is it just 2^7 ?

Thanks a lot ! 🙂

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  1. Editorial Team
    Editorial Team
    2026-05-18T09:15:41+00:00Added an answer on May 18, 2026 at 9:15 am

    How many bits are necessary to address 32 bytes ?

    5 (since 2^5==32). In other words log2(32). If the number in question (32) wasn’t a power of
    two, you would need to round the answer up.

    What are the maximum and minimum unsigned integers that can be represented by 7 bits, 9 bits and 10 bits?

    The minimum unsigned int is always zero, regardless of the number of bits. The maximum unsigned ints are 2^7-1 (127), 2^9-1 (511) and 2^10-1 (1023) respectively.

    This link provides a decent enough explanation.

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