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18 May, 21:42

Consider a system that contains 32K bytes. Assume we are using byte addressing, that is assume that each byte will need to have its own address, and therefore we will need 32K different addresses. For convenience, all addresses will have the same number n, of bits, and n should be as small as possible. What is the value of n?

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  1. 18 May, 22:07
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    n = 15

    Explanation:

    Using

    Given

    Number of bytes = 32k

    At the smallest scale in the computer, information is stored as bits and bytes.

    In the cases when used to describe data storage bits/bytes are calculated as follows:

    Number of bytes when n = 1 is 2¹ = 2 bytes

    When n = 2, number of bytes = 2² = 4 bytes

    n = 3, number of bytes = 2³ = 8 bytes

    So, the general formula is

    Number of bytes = 2^n

    In this case

    2^n = 32k

    Note that 1k = 1024 bytes,

    So, 32k = 32 * 1024 bytes

    Thus, 2^n = 32 * 1024

    2^n = 2^5 * 2^10

    2^n = 2^15

    n = 15.
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