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Consider a system which supports only 1-address type instructions. The size of memory the system has is $2^m$ KB. The system supports ' i ' distinct instructions. The length of an instruction is ____ Bytes

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mention source of the question in title.
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The reason behind this is that 1-address type instruction contains two-part opcode and operand.

opcode will require a decoded form of how many instructions were given ie. $\log_2i$

operand will be decoded form of size of memory ie. $log_2({2^m} KB) = \log_2(2^{m + 10}) = m + 10$ 

Therefore, length of instruction is $\log_2i + m + 10$ bits $=$ $\frac {(\log_2i + m + 10)}{8}$ bytes

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