GATE CS 2006 Q09 — Machine Instructions and Addressing Modes

Assuming the memory is byte-addressable (which is the standard default for such problems):

• The instruction size is $24\text{ bits} = \frac{24}{8} = 3\text{ bytes}$.
• Since each instruction is $3\text{ bytes}$ long, the Program Counter (PC) must advance by multiples of $3$ bytes for each subsequent instruction.

Given that the program starts at address $300$, any legal PC value must be of the form:

This implies that $(\text{PC} - 300)$ must be perfectly divisible by $3$. Let's test the given options:

• Option A (400): $400 - 300 = 100$, which is not divisible by $3$.
• Option B (500): $500 - 300 = 200$, which is not divisible by $3$.
• Option C (600): $600 - 300 = 300$, which is divisible by $3$ ($300 = 3 \times 100$). This is a legal PC address.
• Option D (700): $700 - 300 = 400$, which is not divisible by $3$.

Hence, 600 is the only legal program counter value among the choices.