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A CPU has $24$-$bit$ instructions. A program starts at address $300$ (in decimal). Which one of the following is a legal program counter (all values in decimal)?

1. $400$
2. $500$
3. $600$
4. $700$

Option $(C)$. $24$ bits = $3$ bytes instructions. So, PC will have multiples of $3$ in it.
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CPU is byte addressable always if not given
bit addressable naya aya hh kya market me? 😁

There is nothing as bit addressable!

C) is the ans

size of instruction  =  24/8 = 3 bytes.

Program Counter can shift 3 bytes at a time to jump to next instruction.

So the given options must be divisible by 3. only 600 is satisfied.

Can you explain  why you divided 24 by 8?
Converting bit to byte and as a instruction size is 3 bytes each,program counter will shift 3 bytes each time so address will be multiple of 3.
After each instruction cycle the PC needs to be updated to point to the next instruction in memory.

INSTRUCTION SIZE : 24BITS or 3Bytes

Address of next instruction : 303 then 306 and so on.

So it is an AP series and only 600 comes under this series. So answer is C

Actually the question is incomplete. It should be mentioned in the question that the

CPU is byte-addressible so that the PC will increment byte by byte. Now because the instruction is 3byte, the PC has to increment 3 times.

So the answer is 600, since it is the only option which is multiple of 3.

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